poj 2192 Zipper DP

dp[i][j]表示s1[]的前i个字符与s2[]的前j个字符能否成功组合，值分别为0，1

如果dp[i-1][j] == 1并且s1[i-1] == s3[i-1+j]（s1,s2,s3都是从0开始的，所以取其中某个字符时要减1），则dp[i][j] = 1 ; 

如果dp[i][j-1] == 1并且s2[i][j-1] == s3[i+j-1]，则dp[i][j] = 1;

#include<iostream>#include<cstring>using namespace std;int main(){int n,k,dp[210][210];char s1[210],s2[210],s3[410];scanf("%d",&n);for(k = 1;k <= n;k++){scanf("%s%s%s",s1,s2,s3);memset(dp,0,sizeof(dp));dp[0][0] = 1;int len1 = strlen(s1);int len2 = strlen(s2);int i,j;for(i = 0;i <= len1;i++)for(j = 0;j <= len2;j++){if(i > 0&&dp[i-1][j]&&s3[i+j-1] == s1[i-1])dp[i][j] = 1;if(j > 0&&dp[i][j-1]&&s3[i+j-1] == s2[j-1])dp[i][j] = 1;}printf("Data set %d: ",k);if(dp[len1][len2])printf("yes\n");elseprintf("no\n");}return 0;}