hdu 1501 简单dp （最长公共子串的变形）

没啥好说的，就是公共子串的一种变形，状态定义基本一致，

就是dp[i][j],分别表示当前状态已经匹配到两个串各自的位置，转移时，看和最终串的匹配结果

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX 507

using namespace std;

int t;
char s[MAX] , s1[MAX] , s2[MAX];
bool dp[MAX][MAX];


int main ( )
{
    scanf ( "%d" , &t );
    int c = 1;
    while ( t-- )
    {
        scanf ( "%s" , s1+1 );
        scanf ( "%s" , s2+1 );
        scanf ( "%s" , s+1 );
        int len1 = strlen ( s1+1 );
        int len2 = strlen ( s2+1 );
        int len = strlen ( s+1 );
        memset ( dp , 0 , sizeof ( dp ) );
        dp[0][0] = 1;
        for ( int i = 1 ; i <= len ; i++ )
            for ( int j = 0 ; j <= len1 ; j++ )
            {
                if ( j > i || i - j > len2 ) continue;
                if ( s1[j] == s[i] && i-j >=0 && j > 0 &&  dp[j-1][i-j] )
                    dp[j][i-j] = 1;
                if ( s2[i-j]==s[i] && i-j-1 >= 0 &&dp[j][i-j-1] )
                    dp[j][i-j] = 1;
            }
        printf ( "Data set %d: " , c++ );
        if ( dp[len1][len2] ) printf ( "yes\n" );
        else printf ( "no\n" );
    }
}