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 + ;
 bool solve(string s1, string s2, string s)
 {
     bool dp[maxn][maxn] = {};
     for (int i = ; i <= s1.size(); ++i)
         if (s1[i - ] == s[i - ])
             dp[i][] = ;
     for (int i = ; i <= s2.size(); ++i)
         if (s2[i - ] == s[i - ])
             dp[][i] = ;
     for (int i = ; i <= s1.size(); ++i)
         for (int j = ; j <= s2.size(); ++j)
         {
             if ((s1[i - ] == s[i + j - ] && dp[i - ][j]) ||
                     (s2[j - ] == s[i + j - ] && dp[i][j - ]))
                 dp[i][j] = ;
         }
     return dp[s1.size()][s2.size()];
 }
 int main()
 {
     int T;
     cin >> T;
     for (int i = ; 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 ;
 }