poj 2192 (DP)

这个题题目意思是给你三个字符串str1，str2，str3.将str3从左自右扫描，去匹配str1和str2中的元素，不可重复，若存在一种匹配方法使得str1和str2都被匹配完全了，则输出yes，否则no　　

用布尔变量dp[i][j]表示组成一个字符串用了第一个字符串的i个字符,用了第二个字符串的j个字符,那么题目所要求的就是dp[len1][len2].

// File Name: 2192.cpp
// Author: Missa_Chen
// Created Time: 2013年07月08日 星期一 09时09分47秒

#include <iostream>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <set>
#include <cstdlib>
#include <vector>
#include <time.h>

using namespace std;

#define LL long long
const int maxn = 2e2 + 5;
bool solve(string s1, string s2, string s)
{
    bool dp[maxn][maxn] = {0};
    for (int i = 1; i <= s1.size(); ++i)
        if (s1[i - 1] == s[i - 1]) 
            dp[i][0] = 1;
    for (int i = 1; i <= s2.size(); ++i)
        if (s2[i - 1] == s[i - 1]) 
            dp[0][i] = 1;
    for (int i = 1; i <= s1.size(); ++i)
        for (int j = 1; j <= s2.size(); ++j)
        {
            if ((s1[i - 1] == s[i + j - 1] && dp[i - 1][j]) || 
                    (s2[j - 1] == s[i + j - 1] && dp[i][j - 1]))
                dp[i][j] = 1;
        }
    return dp[s1.size()][s2.size()];
}
int main()
{
    int T;
    cin >> T;
    for (int i = 1; i <= T; ++i)
    {
        string s1, s2, s;
        cin >> s1 >> s2 >> s;
        cout << "Data set "<< i <<": ";
        if (solve(s1, s2, s)) cout <<"yes" << endl;
        else cout << "no" << endl;
    }
    return 0;
}