POJ 2192 Zipper

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题目地址：http://poj.org/problem?id=2192

思路：因为只能依次取，所以c[i+j]这个字符肯定是由a[i]或者b[j]构成的

AC代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>const int inf = 0x3f3f3f3f;//1061109567typedef long long LL;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;char a[210],b[210],c[420];int dp[210][210];int main(){    int cas = 1,t;    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        scanf("%s%s%s",a+1,b+1,c+1);        int len1 = strlen(a+1);        int len2 = strlen(b+1);        dp[0][0] = 1;        for(int i=0; i<=len1; i++)        {            for(int j=0; j<=len2; j++)            {                if(i >= 1 && c[i+j] == a[i] && dp[i-1][j])                    dp[i][j] = 1;                if(j >= 1 && c[i+j] == b[j] && dp[i][j-1])                    dp[i][j] = 1;            }        }        if(dp[len1][len2])            printf("Data set %d: yes\n",cas++);        else            printf("Data set %d: no\n",cas++);    }    return 0;}

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