hdu5707-Combine String（DP）

Problem Description

Given three strings a, b and c , your mission is to check whether c is the combine string of a and b .
A string c is said to be the combine string of a and b if and only if c can be broken into two subsequences, when you read them as a string, one equals to a , and the other equals to b .
For example, ``adebcf'' is a combine string of ``abc'' and ``def''.

Input

Input file contains several test cases (no more than 20). Process to the end of file.
Each test case contains three strings a, b and c (the length of each string is between 1 and 2000).

Output

For each test case, print ``Yes'', if c is a combine string of a and b , otherwise print ``No''.

Sample Input

abc

def

adebcf

abc

def

abecdf

Sample Output

Yes

No

思路：dp[i][j]表示第一个字符串用了前i个位置（第i个位置已匹配），第二个字符串的前j个位置（第j个位置已匹配）
是否可以对c串成功匹配（成功匹配则必然会匹配到c串的前i+j个位置）。
dp[i][j]==1则表示可以成功匹配
dp[i][j]==0则表示无法成功匹配

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
const int maxn=4005;
int dp[maxn][maxn];
int main()
{
    string a,b,c;
    while(cin>>a>>b>>c){
        memset(dp,0,sizeof(dp));
        if(c.size()!=a.size()+b.size())
            printf("No\n");
        else{
            dp[0][0]=1;
            for(int i=0;i<=a.size();i++){
                for(int j=0;j<=b.size();j++){
                    if(a[i]==c[i+j]){
                        dp[i+1][j]|=dp[i][j];
                    }
                    if(b[j]==c[i+j]){
                        dp[i][j+1]|=dp[i][j];
                    }
                }
            }
            if(dp[a.size()][b.size()]==1){
                printf("Yes\n");
            }else{
                printf("No\n");
            }
        }
    }
    return 0;
}