USACO 4.3 analysis

 Topic 1: Buy low, buy lower

** Description: How many different sequences are in the longest descending subsequence?
** Algorithm: DP + BIGN
** Analysis: BASIC b[i] = b[j]+1 | a[j]> a[i] and b[j]+1> b[i], 1 <= j <i
    If the calculation is repeated, then
       cnt[i] += cnt[j] | a[j]> a[i] and b[j]+1 = b[i]. 1 <= j <i;
       cnt[i ] = MAX(1);
** Special Test: 9 7 5 8 5 1 Dp 1 2 3 2 3 4 Num 1 1 1 1 2 2
 For the first 5 and the second 5, their dp is the same. But num is different.
 Here we can take the last 5 for the following reasons: For a sequence that is not the last 5, the last 5 must be available.
 For example, for the second 5, the 9 75 sequence of the first 5, the second 5 can also be obtained. And the following 5 may have more options,
 such as the second 5 in the above example, you can also get the 9 8 5 sequence. So here, the num corresponding to the last 1 should be 2.
#include <stdio.h> #include <string.h> #define MAXN 5010 #define BASE 100000000 #define MAXL 100 struct BIGN { int L; int a[MAXL]; BIGN() { L = 1; memset(a,0,sizeof(a)); } void display() { printf("%d", a[L]); for (int i = L-1; i >= 1; i--) printf("%08d", a[i]); } void plus (BIGN c) { if (c.L > L) L = c.L; for (int i = 1; i <= L; i++) { a[i] += c.a[i]; if (a[i] >= BASE) { a[i+1] += a[i] / BASE; a[i] %= BASE; } } if (a[L+1]) L++; } }; int n, i, j; int a[MAXN], b[MAXN]; BIGN cnt[MAXN]; int main(){ freopen("buylow.in", "r", stdin); freopen("buylow.out", "w", stdout); scanf("%d", &n); for (i = 1; i <= n+1; i++) { if (i <= n) scanf("%d", &a[i]); else a[i] = 0; b[i] = 1; bool flag = false; for (j = i-1; j >= 1; j--) if (a[j] > a[i] && b[j]+1 > b[i]) b[i] = b[j]+1; int y = -1; for (j = i-1; j >= 1; j--) /*前两个条件为递推的累加，后一个为去掉重复(a[j]==j,只取最近一个)*/ if (a[j] > a[i] && b[j]+1 == b[i] && (y < 0 || a[j] < y)) { y = a[j]; cnt[i].plus(cnt[j]); flag = true; } if (!flag) cnt[i].a[1] = 1; } printf("%d ", b[n+1]-1); cnt[n+1].display(); printf("/n"); return 0; }

 

Topic 2: The Primes

A very troublesome topic, it took a long time to stop TLE, and I don’t want to write it a second time!

/* ID: zhangji42 PROG: prime3 LANG: C++ */ #include <iostream> #include <vector> #include <algorithm> #include <string> #include <string.h> #include <stdio.h> #include <sstream> #include <time.h> using namespace std; const int maxp = 100000; struct type { int num, n[6]; }; bool h[maxp] = {0}; bool pri[10][10][10][10][10]={0}; int len = 0, sum, first; string s[200]; int f, mid; type n1, n2, n3, n4, n5, n6; type type1[10][100];//means all the primes and all odd digits type type2[10][100];//not zero type type4[10][100]; type type3[10][10][10][100]; int main() { freopen("prime3.in", "r" ,stdin); freopen("prime3.out", "w", stdout); void makeprime(); void work(); int get(int,int); scanf("%d%d", &sum, &first); makeprime(); work(); sort(s, s+len); for (int i = 0; i < len; i++) { if (i) cout << endl; cout << s[i]; } return 0; } int get(int n, int k) { if (k == 1) return n/10000; if (k == 2) return (n/1000) % 10; if (k == 3) return (n/100) % 10; if (k == 4) return (n/10) % 10; if (k == 5) return n % 10; } void makeprime() { for (int i = 2; i < maxp; i++) if (!h[i]) { for (int j = i+i; j < maxp; j += i) h[j] = 1; if (i <= 10000) continue; bool odd = true, zero = false; type tmp; tmp.num = i; int ssum = 0, ii = i; for (int j = 1; j <= 5; j++) { tmp.n[j] = get(i, j); if (tmp.n[j] % 2== 0) odd = false; if (!tmp.n[j]) zero = true; ssum += ii % 10; ii /= 10; } if (ssum != sum) { h[i] = 1; continue; } pri[get(i, 1)][get(i, 2)][get(i, 3)][get(i, 4)][get(i, 5)] = 1; //printf("%d/n", i); if (tmp.n[1] == first) { type4[tmp.n[5]][++type4[tmp.n[5]][0].num] = tmp; if (!zero) type2[tmp.n[5]][++type2[tmp.n[5]][0].num] = tmp; } if (odd) type1[tmp.n[5]][++type1[tmp.n[5]][0].num] = tmp; type3[tmp.n[1]][tmp.n[3]][tmp.n[5]][++type3[tmp.n[1]][tmp.n[3]][tmp.n[5]][0].num] = tmp; } } void work() { int i, j, k, i1, i2, i3, ff, ii, jj; int left1, left2, left3, left4, row2, row3, row4, col2, col3, col4; for (ff = 0; ff < 5; ff++) { f = ff*2+1; for (i = 1; i <= type1[f][0].num; i++) for (j = 1; j <= type1[f][0].num; j++) { n1 = type1[f][i]; n2 = type1[f][j]; ii = n1.n[1];jj = n2.n[1]; for (i2 = 1; i2 <= type4[f][0].num; i2++) { n4 = type4[f][i2]; mid = n4.n[3]; for (k = 1; k <= type3[ii][mid][jj][0].num; k++) { n6 = type3[ii][mid][jj][k]; for (i1 = 1; i1 <= type2[ii][0].num; i1++) { n3 = type2[ii][i1]; left1 = sum - n3.n[2] - n4.n[2] - n6.n[4] - n2.n[2]; if (left1 < 0 || left1 >= 10) continue; left3 = sum - n3.n[4] - n6.n[2] - n4.n[4] - n2.n[4]; if (left3 < 0 || left3 >= 10) continue; if (!pri[n3.n[2]][n4.n[2]][left1][n6.n[4]][n2.n[2]]) continue;//row2 if (!pri[n3.n[4]][n6.n[2]][left3][n4.n[4]][n2.n[4]]) continue;//row4 for (i3 = 1; i3 <= type2[jj][0].num; i3++) { n5 = type2[jj][i3]; left2 = sum - n5.n[2] - n4.n[2] - n6.n[2] - n1.n[2]; if (left2 < 0 || left2 >= 10) continue; left4 = sum - n5.n[4] - n6.n[4] - n4.n[4] - n1.n[4]; if (left4 < 0 || left4 >= 10) continue; if (!pri[n3.n[3]][left2][n4.n[3]][left4][n2.n[3]]) continue;//row3 if (!pri[n5.n[2]][n4.n[2]][left2][n6.n[2]][n1.n[2]]) continue;//col2 if (!pri[n5.n[3]][left1][n4.n[3]][left3][n1.n[3]]) continue;//col3 if (!pri[n5.n[4]][n6.n[4]][left4][n4.n[4]][n1.n[4]]) continue;//col4 /*把方案整个保存到一个字符串中，用了字符串文件流*/ ostringstream ss; ss<<n5.num<<'/n' <<n3.n[2]<<n4.n[2]<<left1<<n6.n[4]<<n2.n[2]<<'/n' <<n3.n[3]<<left2<<n4.n[3]<<left4<<n2.n[3]<<'/n' <<n3.n[4]<<n6.n[2]<<left3<<n4.n[4]<<n2.n[4]<<'/n' <<n1.num<<'/n'; s[len]=ss.str(); len++; } } } } } } } /* Test 1: TEST OK [0.011 secs, 6040 KB] Test 2: TEST OK [0.011 secs, 6040 KB] Test 3: TEST OK [0.022 secs, 6040 KB] Test 4: TEST OK [0.043 secs, 6040 KB] Test 5: TEST OK [0.032 secs, 6040 KB] Test 6: TEST OK [0.076 secs, 6040 KB] Test 7: TEST OK [0.076 secs, 6040 KB] Test 8: TEST OK [0.130 secs, 6040 KB] Test 9: TEST OK [0.108 secs, 6040 KB] Test 10: TEST OK [0.184 secs, 6040 KB] */

 

Topic 3: Street Race

** Description: The first question is to find the cut of a directed graph, and the second is to find the cut of an undirected graph.
** Algorithm: DFS
** Analysis: 3 properties of a complete graph:
    1. Any point can be reached from the starting point ;
    2. can reach any point end;
    3. end a is 0;
    first question guarantees that no answer from the division of the 1 to 2 divided side;
    the second question answer 2 while ensuring not to split the split 1 Side;
   so the second question can be developed on the basis of the first question. At the same time, it can be guaranteed that the two parts of the split are complete graphs!
    However, it seems that there is no judgment in the data that the segmentation point of the second question has no self-loop edges.

I found this kind of data when I handed it in, but it is calculated according to what is possible

/* ID: zhangji42 TASK: race3 LANG: C++ */ #include <stdio.h> #include <string.h> //#include <vector> #include <iostream> #define MAXN 60 using namespace std; int f[MAXN][MAXN], vis[MAXN]; //vector <int> ans1; //vector <int> ans2; int ans1[MAXN]={0}; int s1 = 0; int ans2[MAXN]={0}; int s2 = 0; int i, n; bool DFS(int x) { if (x == n) return true; for (int i = 1; i <= n; i++) if (!vis[i] && f[x][i]) { vis[i] = 1; if (DFS(i)) return true; } return false; } bool IsCut(int x) { for (int i = 0; i <= n; i++) if (f[x][i]) { if (vis[i] == 1) return true; else if(vis[i] != 2) { vis[i] = 2; if (IsCut(i)) return true; } } return false; } int main(){ freopen("race3.in", "r", stdin); freopen("race3.out", "w", stdout); n = 0; do { int _u; scanf("%d", &_u); if (_u == -1) break; while (_u != -2) { f[n][_u] = 1; // if (n == _u) ans2[n] = -1; scanf("%d", &_u); } n++; } while (1); n--; memset(vis,0,sizeof(vis)); for (i = 1; i < n; i++) { memset(vis,0,sizeof(vis)); vis[0] = 1; vis[i] = 2; if (!DFS(0)) { ans1[i] = 1, s1++; if (ans2[i] >= 0 && !IsCut(i)) ans2[i] = 1, s2++; } } printf("%d", s1); for (i = 1; i <= n ; i++) if (ans1[i]==1) printf(" %d", i); printf("/n%d", s2); for (i = 1; i <= n; i++) if (ans2[i]==1) printf(" %d", i); printf("/n"); return 0; } /********************************************** ** Description: 第一问为求有向图的割，第二问为求无向图的割 ** Algorithm: DFS ** Analysis: 完全图的3个性质： 1.任意一个点从起点可达； 2.任意一个点可到达终点； 3.终点出度为0； 第一问的答案保证了没有从分割1到分割2的边； 第二问的答案同时要保证没有分割2到分割1的边； 故第二问可以在第一问的基础上展开。同时也可以保证分割的两部分都是完全图！ 但是貌似数据里面没有判断第二问的分割点没有自环边。 ***********************************************/

Topic 4: Letter Game

** Description: The key to this question is to understand the meaning of the question:
** The correct understanding should be to find a combination of one or two words in the given DICT, so
      that the number of occurrences of each letter is not more than the given input The number of each letter in the string
** Analysis: At first, I might be afraid of enumeration. It should feel that the data range is large, but after preliminary selection, only a small part of the
    two cases can be enumerated. Just one click
** NOTE: Since it needs to be in lexicographical order, the last one can be set as an empty string
#include <algorithm> #include <stdio.h> #include <string.h> #include <iostream> #define MAXM 1000 using namespace std; const int v[26] = {2,5,4,4,1,6,5,5,1,7,6,3,5,2,3,5,7,2,1,2,4,6,6,7,5,7}; char s[10], word[MAXM][10]; int t[26], pt[26], f[MAXM][MAXM], d[MAXM]; int i, j; int check(int i, int j) { memcpy(pt,t,sizeof(t)); int p; for (p = 0; word[i][p]; p++) if (--pt[word[i][p]-'a'] < 0) return -1; for (p = 0; word[j][p]; p++) if (--pt[word[j][p]-'a'] < 0) return -1; return d[i]+d[j]; } int main(){ FILE *fin = fopen("lgame.in", "r"); FILE *dict = fopen("lgame.dict", "r"); FILE *fout = fopen("lgame.out", "w"); fscanf(fin,"%s", s); memset(t,0,sizeof(t)); for (i = 0; s[i]; i++) t[s[i]-'a']++; int ans = 0, m = 0, sum = 0; while (fscanf(dict,"%s", s) && s[0] != '.') { memcpy(pt,t,sizeof(t)); sum = 0; for (i = 0; s[i]; i++) { if (--pt[s[i]-'a'] < 0) break; sum += v[s[i]-'a']; } if (s[i]) continue; strcpy(word[m], s); d[m++] = sum; } word[m][0] = '/0'; d[m++] = 0; for (i = 0; i < m-1; i++) for (j = i+1; j < m; j++) { f[i][j] = check(i, j); if (f[i][j] > ans) ans = f[i][j]; } fprintf(fout, "%d/n", ans); for (i = 0; i < m-1; i++) for (j = i+1; j < m; j++) if (f[i][j] == ans) if (j == m-1) fprintf(fout, "%s/n", word[i]); else fprintf(fout, "%s %s/n", word[i], word[j]); return 0; }