机试-搜索专题
深搜
堆和栈的区别：
https://www.cnblogs.com/myblesh/archive/2012/03/14/2396409.html
定义全局变量或者自己malloc，栈的空间是有限的。
背包问题
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 using namespace std; 12 // 输出背包问题的最佳方案 13 const int maxn = 20000; 14 int n,v; 15 int w[maxn]; 16 int c[maxn]; 17 vector<int> tmp,ans; 18 int maxc; 19 void dfs(int i,int sumw,int sumc) 20 { 21 if (i == n) 22 { 23 if (sumc > maxc) 24 { 25 maxc = sumc; 26 ans = tmp; // the way of vector assign 27 } 28 return ; 29 } 30 if (sumw+w[i] <= v) 31 { 32 tmp.push_back(i); 33 dfs(i+1,sumw+w[i],sumc+c[i]); 34 tmp.pop_back(); 35 } 36 dfs(i+1,sumw,sumc); 37 } 38 int main() 39 { 40 while (cin>>n>>v) 41 { 42 for (int i=0;i<n;i++) 43 cin >> w[i]; 44 for (int i=0;i<n;i++) 45 cin >> c[i]; 46 dfs(0,0,0); 47 for (int i=0;i<ans.size();i++) 48 cout << ans[i] << ' '; 49 cout << endl; 50 cout << maxc << endl; 51 52 } 53 54 55 return 0 ; 56 }
N个数选K个使平方和最大且和为X
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 using namespace std; 12 // 从N个数中选择K个数，使和为x的基础上平方和最大 13 const int maxn = 2000; 14 int n,k,x; 15 int maxsum; 16 vector<int> tmp,ans; 17 int a[maxn]; 18 void dfs(int i,int num,int sum,int psum) 19 { 20 // i 代表第i个被选择的数,num代表一共挑选的数的个数，sum为其和，psum为平方和 21 if (i==n) 22 { 23 if (psum > maxsum && num == k && sum == x) 24 { 25 maxsum = psum; 26 ans = tmp; 27 } 28 return ; 29 } 30 dfs(i+1,num,sum,psum); 31 if (sum+a[i]<=x) 32 { 33 tmp.push_back(a[i]); 34 dfs(i+1,num+1,sum+a[i],psum+a[i]*a[i]); 35 tmp.pop_back(); 36 } 37 } 38 int main() 39 { 40 while (cin>>n>>k>>x) 41 { 42 maxsum = 0; 43 for (int i=0;i<n;i++) 44 cin >> a[i]; 45 dfs(0,0,0,0); 46 for (int i=0;i<ans.size();i++) 47 cout << ans[i] << ' '; 48 cout << endl; 49 cout << maxsum << endl; 50 } 51 52 53 return 0 ; 54 } 55 // 1.127s 56 void dfs(int i,int num,int sum,int psum) 57 { 58 // K个并且和为X时才判断 59 if (num == k && sum == x) 60 { 61 if (sum > maxsum) 62 { 63 maxsum = sum; 64 ans = tmp; 65 } 66 return ; 67 } 68 // 其他情况直接return 69 if (i>n-1 || num > k || sum>x) 70 return ; 71 if (a[i]+sum <= x) 72 { 73 tmp.push_back(i); 74 dfs(i+1,num+1,sum+a[i],psum+a[i]*a[i]); 75 tmp.pop_back(); 76 } 77 dfs(i+1,num,sum,psum); 78 }
codeforces round 479 D divide by three, multiple by two
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 using namespace std; 12 13 14 int n; 15 long long a[101]; 16 bool visited[101]; 17 long long f[101]; 18 19 bool dfs(int &k) 20 { 21 if (k==n) 22 return true; 23 long long x = f[k-1]; 24 for (int i=0;i<n;i++) 25 { 26 if (visited[i]) 27 continue; 28 if ((a[i] == x/3 && (x%3==0)) || a[i] == x*2) 29 { 30 f[k++] = a[i]; 31 visited[i] = true; 32 if (dfs(k)) return true; 33 visited[i] = false; 34 k--; 35 } 36 } 37 return false; 38 } 39 int main() 40 { 41 42 int k; 43 while (cin>>n) 44 { 45 for (int i=0;i<n;i++) 46 { 47 cin >> a[i];visited[i] = false; 48 } 49 k = 0; 50 for(int i=0;i<n;i++) 51 { 52 visited[i] = true; 53 f[k++] = a[i]; 54 if (dfs(k)) break; // 只要找到一组就直接break，不用继续循环了 55 visited[i] = false; 56 k--; 57 } 58 for (int i=0;i<n;i++) 59 cout << f[i] <<' '; 60 cout << endl; 61 62 63 64 } 65 66 return 0 ; 67 }
return true / false / break 是为了在找到正确答案后立即输出而不是继续循环到末尾
解法2
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 using namespace std; 12 13 // divide by three, multiple by two 14 int n; 15 long long a[101];// 8B 2^64 = 1024^6 = e^18 16 vector<long long >ans; 17 bool flag,visited[101] ; 18 void dfs(int i,int sum) // i: the i th number ; sum:selected number 19 { 20 if (sum==n) 21 { 22 for (int i=0;i<n-1;i++) 23 cout << ans[i] << ' '; 24 cout << ans[n-1]<<endl; 25 flag = true; 26 return ; 27 } 28 29 //4 8 6 3 12 9 30 for (int j=0;j<n;j++) 31 { 32 if (i!=j && !visited[j]) 33 { 34 if ((a[j]==a[i]/3 && a[i]%3==0) || a[j] == a[i]*2 ) 35 { 36 ans.push_back(a[j]);visited[j]=true; 37 dfs(j,sum+1); 38 if (!flag) 39 { 40 ans.pop_back();visited[j] = false; 41 } 42 } 43 } 44 } 45 } 46 int main() 47 { 48 while (cin>>n) 49 { 50 for (int i=0;i<n;i++) 51 { 52 cin >> a[i];visited[i]=false; 53 } 54 55 ans.clear(); 56 flag = false; 57 for (int i=0;i<n;i++) 58 { 59 ans.push_back(a[i]);visited[i]=true; 60 dfs(i,1); 61 ans.pop_back();visited[i] = false; 62 } 63 64 } 65 66 67 68 return 0 ; 69 }
全排列
- 10 min
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 using namespace std; 12 // 全排列 13 int n,sum; 14 bool visited[101]; 15 vector<int> ans; 16 void dfs(int sum) 17 { 18 if (sum == n) 19 { 20 for (int i=0;i<n-1;i++) 21 cout << ans[i] <<" "; 22 cout << ans[n-1] << endl; 23 return; 24 } 25 for (int i=1;i<=n;i++) 26 { 27 if (!visited[i]) 28 { 29 visited[i]= true; 30 ans.push_back(i); 31 dfs(sum+1); 32 visited[i]= false; 33 ans.pop_back(); 34 } 35 } 36 } 37 int main() 38 { 39 while (cin>>n) 40 { 41 for (int i=1;i<=n;i++) 42 { 43 visited[i] = false; 44 } 45 ans.clear(); 46 dfs(0); // 选择0个元素 47 } 48 49 50 return 0 ; 51 }
组合 递归
- 15min
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 using namespace std; 12 // 递归 求组合 13 int n,r,num; // num:selected numbers 14 vector<int > ans; 15 void dfs(int i,int num) 16 { 17 // 判断第I个数是否要被选择 18 if (i == n) 19 { 20 if (num == r) 21 { 22 for (int i=0;i<r-1;i++) 23 cout << ans[i] <<" "; 24 cout << ans[r-1] << endl; 25 } 26 return; 27 } 28 ans.push_back(i+1); 29 dfs(i+1,num+1); 30 ans.pop_back(); 31 32 dfs(i+1,num); 33 } 34 int main() 35 { 36 while (cin>>n>>r) 37 { 38 ans.clear(); 39 dfs(0,0);// 40 } 41 42 43 return 0 ; 44 }
组合 非递归
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <algorithm> 3 #include <iostream> 4 #include <iomanip> 5 #include <cstring> 6 #include <vector> 7 #include <string> 8 #include <queue> 9 #include <stack> 10 #include <map> 11 #include <set> 12 13 using namespace std; 14 15 void permutation(int n, int m) 16 { 17 vector<int> combine; 18 stack<int> buf; 19 bool pop = false; 20 int top; 21 22 buf.push(1); 23 combine.push_back(1); 24 25 while (buf.size()) 26 { 27 if (combine.size() == m) 28 { 29 for (int i = 0; i < m; ++i) 30 { 31 printf("%d", combine[i]); 32 if (i != m - 1) 33 printf(" "); 34 } 35 printf("\n"); 36 pop = true; 37 } 38 39 top = buf.top() + 1; 40 if (top == n + 1) 41 { 42 pop = true; 43 buf.pop(); 44 combine.pop_back(); 45 continue; 46 } 47 48 if (pop) 49 { 50 buf.pop(); 51 combine.pop_back(); 52 pop = false; 53 } 54 55 if (top <= n) 56 { 57 buf.push(top); 58 combine.push_back(top); 59 } 60 } 61 62 } 63 64 int main() 65 { 66 #ifdef _DEBUG 67 freopen("data.txt", "r+", stdin); 68 #endif // _DEBUG 69 70 int n, m, PS = 0; 71 72 while (cin >> n >> m) 73 { 74 permutation(n, m); 75 } 76 77 78 79 return 0; 80 }
组合数+判断素数
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 using namespace std; 12 // 递归 求组合 13 int n,k,num,ans,ans_num; // num:selected numbers 14 int a[21]; 15 bool visited[21]; 16 17 bool isSu(int x) 18 { 19 for (int i=2;i<sqrt(x)+1;i++) 20 { 21 if (x%i == 0) 22 { 23 return false; 24 } 25 } 26 return true; 27 28 } 29 void dfs(int i,int num) 30 { 31 // 判断第I个数是否要被选择 32 if (i == n) 33 { 34 if (num == k && isSu(ans)) 35 { 36 ans_num++; 37 } 38 return; 39 } 40 if (num+1 <= k ) 41 { 42 ans += a[i+1]; 43 dfs(i+1,num+1); 44 ans -= a[i+1]; 45 } 46 dfs(i+1,num); 47 } 48 int main() 49 { 50 while (cin>>n>>k) 51 { 52 for (int i=1;i<=n;i++) 53 { 54 cin >> a[i]; visited[i]= false; 55 } 56 ans = 0;ans_num = 0; 57 dfs(0,0); 58 cout << ans_num << endl; 59 } 60 61 62 return 0 ; 63 }
N皇后
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 using namespace std; 12 // N皇后问题 13 int n,num; 14 bool visited[11]; 15 vector<int> ans; 16 bool flag = false; 17 bool isok(int i,int j) 18 { 19 // 判断第I行皇后能否放在J个列 20 // 1. 不可能同一行，因为I从1-n 21 // 2. 同一列 22 if (visited[j]) 23 return false; 24 // 3. 对角线 25 // i,j / k, l |i-k| = |k-l| 时两个点在对角线上 26 for (int k = 1;k<i;k++) 27 { 28 if (abs(k-i) == abs(ans[k-1]-j)) 29 return false; 30 } 31 return true; 32 } 33 void dfs(int i) 34 { 35 if (i==n) 36 { 37 flag = true; 38 for (int j=0;j<n-1;j++) 39 cout << ans[j] << " "; 40 cout << ans[n-1] << endl; 41 return ; 42 } 43 for (int j=1;j<=n;j++) 44 { // j : column 45 if (isok(i+1,j)) 46 { 47 visited[j] = true; 48 ans.push_back(j); 49 dfs(i+1); 50 visited[j] = false; 51 ans.pop_back(); 52 } 53 54 } 55 } 56 int main() 57 { 58 while (cin>>n) 59 { 60 ans.clear(); 61 for (int i=0;i<=n;i++) 62 visited[i] = false; 63 dfs(0); 64 if (!flag) 65 cout<<"no solute!" <<endl; 66 } 67 68 69 return 0 ; 70 }
迷宫
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 using namespace std; 12 // 走迷宫 13 int n,m; 14 int a[101][16]; 15 struct point 16 { 17 int x,y; 18 }; 19 point s,e,t; 20 21 vector<point> ans; 22 void print() 23 { 24 for (int i=0;i<ans.size()-1;i++) 25 { 26 cout << "("<<ans[i].x<<","<<ans[i].y<<")->"; 27 } 28 cout << "("<<ans[ans.size()-1].x<<","<<ans[ans.size()-1].y<<")"<<endl; 29 } 30 int dir[4][2] = { 31 0,-1, 32 -1,0, 33 0,1, 34 1,0 35 }; 36 void dfs(point p) 37 { // p：当前位置 38 if (p.x == e.x && p.y == e.y) 39 { 40 print(); 41 return; 42 } 43 if (p.x<1 || p.x > n || p.y < 1 || p.y > m) 44 return ; // meet border return 45 if (a[p.x][p.y] == 0) 46 return ; // meet wall return 47 for (int i=0;i<4;i++) 48 { 49 t = p; 50 // 四个方向即四个分支可以深搜 51 t.x = p.x+dir[i][0]; 52 t.y = p.y+dir[i][1]; 53 // 已经访问过的点再往下深搜的过程中不会再次访问，可以设为wall 54 a[p.x][p.y] = 0; 55 ans.push_back(t); 56 dfs(t); 57 a[p.x][p.y] = 1; 58 ans.pop_back(); 59 60 } 61 62 } 63 int main() 64 { 65 66 while (cin>>n>>m) 67 { 68 for (int i=1;i<=n;i++) 69 { 70 for (int j=1;j<=m;j++) 71 { 72 cin >> a[i][j]; 73 } 74 } 75 ans.clear(); 76 cin >> s.x >> s.y; 77 cin >> e.x >> e.y; 78 ans.push_back(s); 79 dfs(s); 80 81 82 } 83 84 85 return 0 ; 86 }
- 本地运行正确但提交不正确
jugs
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 #include <queue> 12 using namespace std; 13 int ca,cb,n; 14 struct node 15 { 16 int x,y; 17 struct node *parent; 18 string word; 19 }s,e,tmp,cur; 20 queue<node> q; 21 vector<string> w; 22 int main() 23 { 24 while (cin>>ca>>cb>>n) 25 { 26 while (!q.empty()) 27 q.pop(); 28 w.clear(); 29 s.x = 0;s.y = 0;s.parent = NULL; 30 e.y = n; 31 q.push(s); 32 while (!q.empty()) 33 { 34 cur = q.front(); 35 q.pop(); 36 // ÅжÏÊÇ·ñΪ×îÖÕ̬ 37 if (cur.y == e.y) 38 { 39 string t = "success"; 40 w.push_back(t); 41 struct node *p;p=&cur; 42 while (p->parent!=NULL) 43 { 44 w.push_back(p->word); 45 p = p->parent; 46 } 47 for (int i = w.size()-1;i>=0;i--) 48 { 49 cout << w[i] << endl; 50 } 51 } 52 53 54 if (cur.x < ca) 55 { 56 tmp = cur; 57 // A is not full 58 tmp.x = ca; 59 tmp.word = "fill A"; 60 tmp.parent = &cur; 61 q.push(tmp); 62 } 63 64 if (cur.y < cb) 65 { 66 tmp = cur; 67 // B is not full 68 tmp.y = cb; 69 tmp.word = "fill B"; 70 tmp.parent = &cur; 71 q.push(tmp); 72 } 73 74 if (cur.x > 0) 75 { 76 tmp = cur; 77 // A is not empty 78 tmp.x = 0; 79 tmp.word = "empty A"; 80 tmp.parent = &cur; 81 q.push(tmp); 82 } 83 84 if (cur.y > 0) 85 { 86 tmp = cur; 87 // B is not empty 88 tmp.y = 0; 89 tmp.word = "empty B"; 90 tmp.parent = &cur; 91 q.push(tmp); 92 } 93 94 if (cur.x > 0 && cur.y < ca) 95 { 96 tmp = cur; 97 if (cur.x >= cb-cur.y) // x more than capacity of B 98 { 99 tmp.y = cb; 100 tmp.x = cur.x - cb + cur.y; 101 } 102 else 103 { 104 tmp.y = cur.x+cur.y; 105 tmp.x = 0; 106 } 107 tmp.word = "pour A B"; 108 tmp.parent = &cur; 109 q.push(tmp); 110 } 111 112 if (cur.y > 0 && cur.x < ca) 113 { 114 tmp = cur; 115 // B have water and A has place to hold 116 if (cur.y >= ca-cur.x) // y more than capacity of A 117 { 118 tmp.x = ca; 119 tmp.y = cur.y - ca + cur.x; 120 } 121 else 122 { 123 tmp.x = cur.x+cur.y; 124 tmp.y = 0; 125 } 126 tmp.parent = &cur; 127 tmp.word = "pour B A"; 128 q.push(tmp); 129 } 130 } 131 } 132 return 0 ; 133 }
- char 型迷宫定义成int型导致爆莫名奇妙的错误。
掉石头迷宫
应该是迷宫状态更新的问题。 1.5小时
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 #include <queue> 12 using namespace std; 13 // 掉石头迷宫问题 14 int t1; 15 char a[9][9],t[9][9]; 16 struct point 17 { 18 int x,y,t; 19 }s,e,tmp,cur; 20 vector<point> stone; 21 queue<point> q; 22 void printa(char a[9][9]) 23 { 24 for (int i=1;i<=8;i++) 25 { 26 for (int j=1;j<=8;j++) 27 cout << a[i][j] << " "; 28 cout << endl; 29 } 30 } 31 void flush(char a[9][9]) // 石头状态刷新一次 32 { 33 /* int x1,y1; 34 for (int i=0;i<stone.size();i++) 35 { 36 x1 = stone[i].x; 37 y1 = stone[i].y; 38 a[x1][y1] = '.'; 39 if (x1+1 <=8) // 超出边界的石头则删除了 40 a[x1+1][y1] = 'S'; 41 } 42 43 */ 44 for (int i=1;i<9;i++) 45 { 46 for (int j=1;j<9;j++) 47 { 48 if (a[i][j] == 'S') 49 { 50 a[i+1][j] = 'S'; 51 a[i][j] = '.'; 52 } 53 } 54 } 55 56 } 57 int dir[9][2] = 58 { 59 -1,0, 60 -1,1, 61 0,1, 62 1,1, 63 1,0, 64 1,-1, 65 0,-1, 66 -1,-1, 67 0,0 68 }; 69 bool judge(point tmp) 70 { 71 if (tmp.x < 1 || tmp.y >8 || tmp.y < 1 || tmp.y > 8) 72 return false; 73 if (a[tmp.x][tmp.y] == 'S' || t[tmp.x][tmp.y] == 'S') // 移动后所在位置有石头或移动后石头下落砸到自己 74 return false; 75 return true; 76 } 77 int main() 78 { 79 cin >> t1; 80 bool flag; 81 while (t1--) 82 { 83 flag = false; 84 // 输入迷宫 85 for (int i=1;i<=8;i++) 86 { 87 for (int j=1;j<=8;j++) 88 { 89 cin >> a[i][j]; t[i][j] = a[i][j]; 90 } 91 } 92 93 /* 存石头位置 94 stone.clear(); 95 for (int i=1;i<9;i++) 96 { 97 for (int j=1;j<9;j++) 98 { 99 if (a[i][j] == 'S') 100 { 101 tmp.x = i;tmp.y = j; 102 stone.push_back(tmp); 103 } 104 } 105 } 106 */ 107 // 清空队列 108 while (!q.empty()) 109 q.pop(); 110 // 初始化队列 111 s.x = 8;s.y = 1;e.x = 1;e.y = 8;s.t = 0; 112 q.push(s); 113 while (!q.empty()) 114 { 115 cur = q.front();q.pop(); 116 if (cur.x == e.x && cur.y == e.y) 117 { 118 flag = true; 119 break; 120 } 121 flush(t); // t是移动后的a 122 printa(t); 123 for (int i=0;i<9;i++) 124 { 125 tmp = cur; 126 tmp.x += dir[i][0]; 127 tmp.y += dir[i][1]; 128 tmp.t ++; 129 if (judge(tmp)) 130 { 131 q.push(tmp); 132 } 133 } 134 flush(a);//a t 同步了 135 printa(a); 136 137 } 138 if (flag) 139 cout << "Yes" << endl; 140 else 141 cout << "No" << endl; 142 getchar(); 143 } 144 return 0 ; 145 }
DP
斐波那契
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 #include <queue> 12 using namespace std; 13 14 int fib(int n) 15 { 16 if (n==1 || n==2) 17 return 1; 18 else 19 return fib(n-1)+fib(n-2); 20 } 21 int main() 22 { 23 int n; 24 while (cin >> n) 25 { 26 cout << fib(n) << endl; 27 } 28 return 0 ; 29 }
- n <= 30 最简单的递归，内存超限： 大量的重复的计算
- 方法：通过空间的损耗来提高计算的速度，一维数组DP记录每个数的数值，未计算过则设置为-1。
- 记忆化搜索 每个数仅算一遍
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 #include <queue> 12 using namespace std; 13 14 int dp[32]; 15 int fib(int n) 16 { 17 if (n==1 || n==2) // 递归边界 18 { 19 return 1; 20 } 21 if (dp[n]!=-1) 22 return dp[n]; // 先判断我们的缓存里面有没有，有的话直接取出否则递归计算 23 else 24 { 25 dp[n] = fib(n-1)+fib(n-2); 26 return dp[n]; 27 } 28 29 } 30 int main() 31 { 32 int n; 33 while (cin >> n) 34 { 35 for (int i=0;i<32;i++) dp[i] = -1; 36 cout << fib(n) << endl; 37 } 38 return 0 ; 39 }
最大连续子序列和
？？？？？？？？？？？？？？？？？ 如何输出最大连续的子序列呢？
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 #include <queue> 12 using namespace std; 13 14 int dp[32]; 15 bool flag[32]; 16 int a[10001]; 17 int main() 18 { 19 int n; 20 while (cin >> n) 21 { 22 for (int i=0;i<n;i++) 23 {cin >> a[i];flag[i] = false;} 24 dp[0] = a[0]; 25 26 for (int i=1;i<n;i++) 27 { 28 dp[i] = max(a[i],dp[i-1]+a[i]); 29 } 30 31 32 33 34 } 35 return 0 ; 36 }
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 #include <queue> 12 using namespace std; 13 14 int dp[32]; 15 bool flag[32][32]; 16 int a[10001]; 17 int main() 18 { 19 int n; 20 while (cin >> n) 21 { 22 for (int i=0;i<n;i++) 23 {cin >> a[i];} 24 for (int i=0;i<32;i++) 25 { 26 for (int j=0;j<32;j++) 27 flag[i][j] = false; 28 } 29 dp[0] = a[0]; flag[0][0] = true; 30 31 for (int i=1;i<n;i++) 32 { 33 dp[i] = max(a[i],dp[i-1]+a[i]); 34 if (dp[i] == a[i]) 35 { 36 flag[i][i] = true; 37 } 38 else 39 { 40 for (int x=0;x<i;x++) 41 flag[i][x] = flag[i-1][x]; 42 flag[i][i] = true; 43 } 44 } 45 int k = 0; 46 int max = dp[0]; 47 for (int i=1;i<n;i++) 48 { 49 if (dp[i] > max) 50 { 51 max = dp[i]; k = i; 52 } 53 } 54 for (int i=0;i<n;i++) 55 { 56 if (flag[k][i]) 57 cout << a[i] <<" "; 58 } 59 60 61 62 63 64 } 65 return 0 ; 66 }
- 运行错误？ 还是格式错误？
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 #include <queue> 12 using namespace std; 13 14 15 int main() 16 { 17 int n; 18 int dp[32]; 19 int a[10001]; 20 while (cin >> n) 21 { 22 for (int i=0;i<n;i++) 23 {cin >> a[i];dp[i] = 1;} 24 25 26 for (int i=1;i<n;i++) 27 { 28 for (int j=0;j<i;j++) 29 { 30 if (a[j] < a[i]) 31 { 32 dp[i] = max(dp[i],dp[j]+1); 33 } 34 } 35 } 36 sort(dp,dp+n); 37 cout << dp[n-1] << endl; 38 39 40 } 41 return 0 ; 42 }
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 #include <queue> 12 using namespace std; 13 14 const int maxn = 200001; 15 int a[maxn]; 16 int n; 17 int dp[maxn]; 18 int main() 19 { 20 while (cin>>n) 21 { 22 for (int i=0;i<n;i++) 23 cin >> a[i]; 24 dp[0] = 1; 25 int ans = 0; // 最大dp值的下标 26 for (int i=1;i<n;i++) 27 { 28 for (int j=0;j<i;j++) 29 { 30 if (a[j] == a[i]-1) 31 { 32 dp[i] = max(dp[i],dp[j]+1); // 找到最大的 33 } 34 } 35 if (dp[i] > dp[ans]) 36 { 37 ans = i; 38 } 39 } 40 cout << dp[ans] << endl; 41 vector<int> p;p.clear(); 42 for (int i = ans,j=0;i>=0;i--) 43 { 44 if (a[i] == a[ans]-j) 45 { 46 p.push_back(i);j++; 47 } 48 49 } 50 for (int i=p.size()-1;i>0;i--) 51 cout << p[i]+1 <<" "; 52 cout << p[0]+1 << endl; 53 54 } 55 56 return 0 ; 57 }
time limited 超时了！
用map存储：
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 #include <queue> 12 using namespace std; 13 14 const int maxn = 200001; 15 int a[maxn]; 16 int n; 17 int dp[maxn]; 18 map<int,int> m; 19 int main() 20 { 21 while (cin>>n) 22 { 23 m.clear(); 24 for (int i=0;i<=n;i++) 25 m[i] = 0; 26 for (int i=1;i<=n;i++) 27 { 28 cin >> a[i]; 29 m[a[i]] = m[a[i]-1]+1; 30 } 31 int ans = 0,k=0; 32 map<int,int>::iterator it; 33 for (it = m.begin();it !=m.end();it++) 34 { 35 if (it->second > ans) 36 { 37 ans = it->second; 38 k = it->first; 39 } 40 } 41 int pos = k - ans + 1; 42 cout << ans << endl; 43 for (int i=1;i<=n&&pos <= k;i++ ) 44 { 45 if (a[i] == pos) 46 { 47 cout << i <<" "; 48 pos++; 49 } 50 } 51 52 53 54 } 55 56 return 0 ; 57 }
非常可乐
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 #include <queue> 12 using namespace std; 13 14 int coco,n,m; 15 struct status 16 { 17 int a,b,c,d; 18 }s,cur,tmp; 19 queue<status> q; 20 const int maxn = 101; 21 int v[maxn][maxn][maxn]; 22 void pour(int a,int b,int c,int d) 23 { 24 if (!v[a][b][c]) 25 { 26 v[a][b][c] = 1; 27 tmp.a = a; 28 tmp.b = b; 29 tmp.c = c; 30 tmp.d = d+1; // 未出现的状态才倒水，并加入队列 31 q.push(tmp); 32 } 33 } 34 int bfs(int a,int b,int c,int d) // d为倒水次数,a,b,c为杯中可乐量 35 { 36 while (!q.empty()) q.pop(); 37 s.a = coco;s.b = s.c = s.d = 0; // 初始状态 38 q.push(s); 39 v[coco][0][0] = 1; // 该状态已入队过，为防止重复入队 40 while (!q.empty()) 41 { 42 cur = q.front();q.pop(); 43 if ((cur.a == coco/2 && cur.b == coco/2 )||(cur.a == coco/2 && cur.c == coco/2 ) || (cur.b == coco/2 && cur.c == coco/2 ) ) 44 return cur.d; 45 //s->n: 46 if (cur.a > 0 && cur.b < n) // a has coco and b is not full 47 pour(cur.a - n + cur.b, n, cur.c, cur.d); 48 //s->m; 49 if ( cur.a > 0 && cur.c < m) 50 pour(cur.a - m + cur.c, cur.b, m, cur.d); 51 //n->s; 52 if (cur.b > 0 && cur.a < coco) 53 pour(cur.a + cur.b, 0, cur.c, cur.d); 54 //m->s; 55 if (cur.c > 0 && cur.a < coco ) 56 pour(cur.a + cur.c, cur.b, 0, cur.d); 57 //n->m 58 if (cur.b > 0 && cur.c < m) 59 { 60 if(cur.b > m - cur.c) 61 pour(cur.a, cur.b - m + cur.c, m, cur.d); 62 else 63 pour(cur.a, 0, cur.b + cur.c, cur.d); 64 } 65 66 //m->n 67 if (cur.c > 0 && cur.b < n) 68 { 69 if(cur.c > n - cur.b) 70 pour(cur.a, n, cur.c - n + cur.b, cur.d); 71 else 72 pour(cur.a, cur.b + cur.c, 0, cur.d); 73 } 74 75 } 76 return 0; 77 } 78 int main() 79 { 80 int ans; 81 while (cin>>coco>>n>>m) 82 { 83 memset(v,0,sizeof(int)*maxn*maxn*maxn); 84 if (coco==0) 85 break; 86 if (coco%2 == 1) 87 { 88 cout<< "NO" << endl;continue; 89 } 90 ans = bfs(coco,0,0,0); 91 if ( ans > 0) 92 { 93 cout << ans << endl; 94 } 95 else 96 cout<< "NO" << endl; 97 } 98 99 100 return 0 ; 101 }
导弹拦截系统：最长不降子序列
1 #include <iostream> 2 #include <stdio.h> 3 #include <stdlib.h> 4 #include <vector> 5 #include <stack> 6 #include <map> 7 #include <string> 8 #include <string.h> 9 #include <algorithm> 10 #include <math.h> 11 #include <queue> 12 using namespace std; 13 14 const int maxn = 1001; 15 int main() 16 { 17 int n; 18 int a[maxn]; 19 int dp[maxn]; // 以I为结尾的不降子序列的长度 20 while (cin>>n) 21 { 22 for (int i=0;i<n;i++) 23 { 24 cin >> a[i]; dp[i] = 1; 25 } 26 int ans = 1; 27 for (int i=1;i<n;i++) 28 { 29 for (int j=0;j<i;j++) 30 { 31 if (a[j] < a[i]) 32 { 33 dp[i] = max(dp[j]+1,dp[i]); 34 } 35 } 36 if (dp[i] > ans) 37 ans = dp[i]; 38 } 39 cout <<ans<< endl; 40 41 42 43 } 44 return 0 ; 45 }