POJ 2192 Zipper

http://poj.org/problem?id=2192

题意：

给出3个字符串，判断第1个和第2个能否组成第3个，要求是每个字符串的相对顺序不能改变。

思路：
d[i][j]表示第1个取第i个字符，第2个取第j个字符时能否组成第3个字符。

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn = 500 + 5;

char s1[maxn], s2[maxn], s3[maxn];
int d[maxn][maxn];

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    int kase = 0;
    while (T--)
    {
        scanf("%s%s%s", s1+1, s2+1, s3+1);
        int len1 = strlen(s1+1);
        int len2 = strlen(s2+1);
        int len3 = strlen(s3+1);
        memset(d, 0, sizeof(d));
        for (int i = 1; i <= len1; i++)
        {
            if (s1[i] == s3[i])
                d[i][0] = 1;
            else break;
        }
        for (int i = 1; i <= len2; i++)
        {
            if (s2[i] == s3[i])
                d[0][i] = 1;
            else break;
        }
        for (int i = 1; i <= len1; i++)
        for (int j = 1; j <= len2; j++)
        {
            if (s1[i] == s3[i + j] && d[i - 1][j] == 1)
                d[i][j] = 1;
            if (s2[j] == s3[i + j] && d[i][j - 1] == 1)
                d[i][j] = 1;
        }
        printf("Data set %d: ", ++kase);
        if (d[len1][len2] == 1)
            printf("yes
");
        else
            printf("no
");
    }
}