第七次集训——常见动态规划模型（清明）

  模型一：石子归并类

题目一：Northeastern Europe 2001, Far-Eastern Subregion

/******************************************************* ** Description: 给定N张卡片，每次只能从中间取卡片， 取走一张卡片的价值为这张卡片的价值和相邻的卡片价值的乘积， 求最后剩下2张卡片时，能获得的最小价值是多少？ ** Analysis： f[i][j] 表示 从 第i张 到 第j张，能或得的最小价值 阶段： 取掉的连续区间的长度 决策： f[i][j] = min(f[i+1][j]+w[i-1]*w[i]*w[j+1], f[i][j-1]+w[j+1]*w[j]*w[i-1], f[i][k-1]+f[k+1][j]+w[k]*w[i-1]*w[j+1]), i < k < j k表示在这个区间中最后取走的牌的编号 ** Time: O(N^3) *******************************************************/ #include <iostream> using namespace std; const int maxn = 101; int a[maxn]; long long f[maxn][maxn]; int n; long long min(long long a, long long b) { if (a < b) return a; else return b; } int main() { scanf("%d", &n); for (int i = 1; i <= n; i++) scanf("%d", &a[i]); for (int i = 2; i < n; i++) f[i][i] = a[i-1]*a[i]*a[i+1]; for (int k = 2; k <= n-2; k++) for (int i = 2; i <= n-k; i++) { int j = i+k-1; f[i][j] = min(f[i+1][j]+a[i-1]*a[i]*a[j+1], f[i][j-1]+a[j+1]*a[j]*a[i-1]); for (int p = i+1; p < j; p++) f[i][j] = min(f[i][j], f[i][p-1]+f[p+1][j]+a[p]*a[i-1]*a[j+1]); } printf("%I64d/n", f[2][n-1]); return 0; }

模型二：同余类

题目一：Northeastern Europe 1999

/***************************************************************** ** Description: 给定N个数，求有这N个数加减能否得到某一个数P的倍数？ ** Analysis： f[i][j] 表示用前i个数能否得到模P余J f[i][j] = f[i-1][(j-w[i])modP] || f[i-1][(j+w[i])modP] 但要注意 % 操作对于负数的情况 -1 % 5 = -1 ** Time: O(P*N) *****************************************************************/ #include <iostream> using namespace std; int n, p, m, state; int h[2][101]; int f(int k) { k = k % p; k = (k+p)%p; return k; } int main() { while (scanf("%d%d", &n, &p) == 2) { memset(h,0,sizeof(h)); state = 0; h[0][0] = 1; for (int i = 0; i < n; i++) { scanf("%d", &m); state = 1 - state; for (int j = 0; j < p; j++) h[state][j] = 0; for (int j = 0; j < p; j++) if (h[1-state][j]) h[state][f(j+m)] = h[state][f(j-m)] = 1; } if (h[state][0]) printf("Divisible/n"); else printf("Not divisible/n"); } return 0; }

模型三：最长不下降子序类

题目一： Waterloo local 2000.09.23

/***************************************************************** ** Description： 按字典序给出N个单词，对一个单词可以进行修改一个字母，添加一个字母，删除一个字母的操作 求出做多选出多少个单词，使得相邻两个满足进行一次操作 ** Analysis： 每次构造出由但前单词可以构造出的所有单词，然后去判断前面是否出现过，找出一个最大值 算法1：trie 算法2：二分 *****************************************************************/ #include <iostream> #include <string.h> using namespace std; const int maxn = 25010; int n, len, k; int f[maxn]; char w[maxn][20], s[20]; int check() { int l = 1, r = n, mid; while (l+1<r) { mid = (l+r)/2; if (strcmp(s, w[mid]) >= 0) l = mid; else r = mid; } if (strcmp(w[l], s) == 0) return l; else return 0; } int main() { freopen("e.in", "r",stdin); freopen("e.out", "w", stdout); n = 0; f[0] = 0; while (scanf("%s", w[++n]) == 1) { len = strlen(w[n]); f[n] = 1; //cout << w[n] << endl; if (n == 1) continue; //changing for (int i = 0; i < len; i++) s[i] = w[n][i]; s[len] = '/0'; for (int i = 0; i < len; i++) { for (char j = 'a'; j < w[n][i]; j++) { s[i] = j; //cout << s << endl; k = check(); if (f[k]+1>f[n]) f[n] = f[k]+1; } s[i] = w[n][i]; } //delete s[len-1] = '/0'; if (len > 1) for (int i = 0; i < len; i++) { for (int j = 0; j < i; j++) s[j] = w[n][j]; for (int j = i+1; j < len; j++) s[j-1] = w[n][j]; //cout << s << endl; len--; k = check(); len++; if (f[k]+1>f[n]) f[n] = f[k]+1; } //addition s[len+1] = '/0'; for (int i = 0; i < len; i++) s[i+1] = w[n][i]; for (int i = 0; i < len; i++) { len++; for (char j = 'a'; j <= w[n][i]; j++) { s[i] = j; //cout << s << endl; k = check(); if (f[k]+1>f[n]) f[n] = f[k]+1; } len--; s[i] = w[n][i]; } } int max = 0; for (int i = 1; i <= n; i++) if (f[i] > max) max = f[i]; printf("%d/n", max); return 0; }

题目二：Northeastern Europe 1998

/*************************************************** ** Desrtiption: 有N个贼，每个贼在某一个时刻Ti来到商店，商店有有0--K的门， 只有当前商店门的编号与贼心中的编号Si附合时才会偷价值为Pi的物品 在一个单位时间内门的编号可以就+1 0 -1 求在可以操作门的情况下，最多能偷走物品数？ ** Analysis： 来的时间有先后，排个序！ f[i] 表示以第i人偷时的最优值，那么 f[i] = max(f[j]+p[i]), |s[i]-s[j]|<=|t[i]-t[j]|, 1 <= j < i ** Time: O(N^2) /*************************************************** #include <iostream> using namespace std; const int maxn = 101; int t[maxn], p[maxn], s[maxn], f[maxn]; int n, k, tt; void swap(int &a, int &b) { int t = a; a = b; b = t; } int main() { freopen("g.in", "r", stdin); scanf("%d%d%d", &n, &k, &tt); for (int i = 1; i <= n; i++) scanf("%d", &t[i]); for (int i = 1; i <= n; i++) scanf("%d", &p[i]); for (int i = 1; i <= n; i++) scanf("%d", &s[i]); //sort for (int i = 1; i < n; i++) for (int j = i+1; j <= n; j++) if (t[j] < t[i]) { swap(t[i], t[j]); swap(p[i], p[j]); swap(s[i], s[j]); } t[0] = 0; s[0] = 0; f[0] = 0; for (int i = 1; i <= n; i++) { f[i] = -1 << 30;//necessary for (int j = 0; j < i; j++) if (abs(s[i]-s[j]) <= abs(t[i]-t[j])) if (f[j] + p[i] > f[i]) f[i] = f[j] + p[i]; } int max = 0; for (int i =1; i <= n; i++) if (f[i] > max) max = f[i]; printf("%d/n", max); return 0; }

模型四：背包类

NUDT-P2180

/****************************************** ** Description: 给定N个物品， 给出这N个物品之间的关系，有直接关系的不能放进同一堆 求把这N个物品放进两堆最小差值是多少？ ** Analysis： 初看好像和背包没有关系？？？对于一个连通分量转化成二分图后，两堆的差值是确定的 所以可以把这个差值作为一个物品的重量去做，转化为M个物品分成两堆的最小差距，就可以用背包完成了 ********************************************/ #include <iostream> using namespace std; const int maxn = 100, maxv = 2010; int n, i, j, now, l, r, sum, c[maxn]; bool d[maxn][maxn], h[maxv], v[maxn]; char map[110]; void dfs(int x, int sgn) { now += sgn*c[x]; v[x] = 1; for (int i = 0; i < n; i++) if (!v[i] && d[x][i]) dfs(i, -sgn); } int main() { freopen("d.in", "r", stdin); while (scanf("%d", &n) == 1) { for (i = 0; i < n; i++) scanf("%d", &c[i]); for (i = 0; i < n; i++) { scanf("%s", map); for (j = 0; j < i; j++) if (map[j] == '1') d[i][j] = d[j][i] = 1; else d[i][j] = d[j][i] = 0; } memset(v,0,sizeof(v)); memset(h,0,sizeof(h)); h[0] = 1; sum = 0; for (i = 0; i < n; i++) if (!v[i]) { now = 0; dfs(i, 1); if (now < 0) now = -now; // cout << now << endl; sum += now; for (j = sum; j >= now; j--) h[j] = h[j] || h[j-now]; } l = sum/2; r = sum-l; while (!h[l]) l--, r++; printf("%d/n", r-l); } return 0; }

模型五：矩阵取数类

题目一：Central Europe 2005

/************************************************************ **　Description: 给定二个矩阵，分别有2中矿藏， 第一种矿藏只能收集到第0列，第二种矿藏只能收集到第0行 一条直线上只能有一种运输方式（从下到上或从右到左） 求获得的最大矿藏 ** Analysis: f[i][j] 表示收集到第i行，第j列的位置时获得的最大矿藏 f[i][j] = max(f[i-1][j]+row[i][j], f[i][j-1]+col[i][j]) *************************************************************/ #include <iostream> using namespace std; const int maxn = 501; int g[maxn][maxn], h[maxn][maxn], f[maxn][maxn]; int n, m, w; int max(int a, int b) { if (a > b) return a; else return b; } int main() { freopen("h.in", "r", stdin); for (scanf("%d%d", &n, &m); n+m>0; scanf("%d%d", &n, &m)) { for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { scanf("%d", &w); g[i][j] = g[i][j-1]+w; }//mine1 for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) { scanf("%d", &w); h[j][i] = h[j][i-1]+w; }//mine2 for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) f[i][j] = max(f[i-1][j]+g[i][j], f[i][j-1]+h[j][i]); printf("%d/n", f[n][m]); } return 0; }

模型六：LCS类

题目一：FZU4月月赛G题

/************************************************* ** Descrition: Given two strings Task One; Get the Largest Common String Task Two: Get the minimal steps to let the string be the same you can insert or delete or change a char ** Algorithm: DP ** Analysis: Denote f[i][j] be the maximal LCS of the first i chars in string1 and first j chars in string2 f[i][j] = max(f[i-1][j], f[i][j-1]) s1[i] != S2[J] f[i-1][j-1]+1 s1[i] == s2[j] Denote g[i][j] be the minimal steps of the first i chars in string1 and first j chars in string2 g[i][j] = min(g[i][j-1], g[i-1][j], g[i-1][j-1])+1, s1[i] != s2[j] g[i-1][j-1] s1[i] == s2[j] **************************************************/ #include <stdio.h> #include <string.h> #define MAXL 1010 int cases, i, j; char s1[MAXL], s2[MAXL]; int f[MAXL][MAXL], g[MAXL][MAXL]; int main() { freopen("g.in","r", stdin); scanf("%d/n", &cases); for (int ci = 1; ci <= cases; ci++) { gets(s1); gets(s2); int n1 = strlen(s1); int n2 = strlen(s2); f[0][0] = f[1][0] = f[0][1] = 0; g[0][0] = 0; g[1][0] = g[1][0] = 1; for (i = 0; i < n1; i++) for (j = 0; j < n2; j++) if (s1[i] == s2[j]) { f[i+1][j+1] = f[i][j]+1; g[i+1][j+1] = g[i][j]; } else { if (f[i+1][j] > f[i][j+1]) { f[i+1][j+1] = f[i+1][j]; } else { f[i+1][j+1] = f[i][j+1]; } if (g[i+1][j] > g[i][j+1]) { g[i+1][j+1] = g[i][j+1]+1; } else { g[i+1][j+1] = g[i+1][j]+1; } if (g[i][j]+1 < g[i+1][j+1]) g[i+1][j+1] = g[i][j]+1; } printf("Case %d: LCS=%d EditStep=%d/n", ci, f[n1][n2], g[n1][n2]); } }