POJ2192 | ZOJ 2401 Zipper

来源：互联网发布：淘宝3c证书编号怎么填编辑：程序博客网时间：2024/05/02 04:57

题意:给出3个字符串A,B,C 判断第3个字符串是否可以通过前2个串穿插变成.

设dp[i][j]表示A的前i个字符串和B的前j个字符串是否能够构成C的前i + j个字符.

则dp[i][j] = max(dp[i - 1][j] * (A[i - 1] == C[i + j - 1]), dp[i][j - 1] * (B[j - 1] == C[i + j - 1])).这里字符串索引从0开始.

base cases:

dp[i][0] = dp[i - 1][0] * (A[i - 1] == C[i])

dp[0][j] = dp[0][j - 1] * (B[j - 1] == C[j])

#include <cstring>#include <string>#include <cstdio>using namespace std;const int MAX = 405;bool dp[MAX][MAX];char A[MAX], B[MAX], C[MAX];int main(int argc, char const *argv[]){int T, caseno = 1;scanf("%d", &T);while(T--){scanf("%s%s%s", A, B, C);memset(dp, false, sizeof(dp));dp[0][0] = 1;int len1 = strlen(A), len2 = strlen(B), len3 = strlen(C);for(int i = 1; i <= len1; ++i){dp[i][0] = dp[i - 1][0] * (A[i - 1] == C[i - 1]);}for(int i = 1; i <= len2; ++i){dp[0][i] = dp[0][i - 1] * (B[i - 1] == C[i - 1]);}for(int i = 1; i <= len1; ++i){for(int j = 1; j <= len2; ++j){dp[i][j] = max(dp[i - 1][j] * (A[i - 1] == C[i + j - 1]), dp[i][j - 1] * (B[j - 1] == C[i + j - 1]));}}if(dp[len1][len2]){printf("Data set %d: yes\n", caseno++);}else{printf("Data set %d: no\n", caseno++);}}return 0;}

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