(字符串模式匹配4.7.10)POJ 2192 Zipper(判断第3个字符串能否有前两个字符串组成)

/*
 * POJ_2192.cpp
 *
 *  Created on: 2013年10月28日
 *      Author: Administrator
 */


#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 210;
int main(){
	int t;
	scanf("%d",&t);
	int  counter = 1;
	char str1[maxn],str2[maxn],str3[maxn*2];//注意,str3所取的长度应该是str1与str2之和
    bool can[maxn][maxn];


	while(t--){
		memset(can,0,sizeof(can));
		can[0][0] = 1;

		scanf("%s%s%s",str1,str2,str3);
	    int k1 = strlen(str1);
	    int k2 = strlen(str2);

	    int i,j;
	    for(i = 0 ; i <= k1 ; ++i){
	    	for(j = 0 ; j <= k2 ; ++j){
	    		if(i >= 1 && str3[i+j-1] == str1[i-1] && can[i-1][j] == 1){//当c[i+j-1] == a[i-1]时,需要看一下c[i+j-1]是否能由a[i-2]和b[j-1]组成
	    			can[i][j] = 1;
	    		}else if(j >= 1 && str3[i+j-1] == str2[j-1] &&  can[i][j-1] == 1){
	    			can[i][j] = 1;
	    		}
	    	}
	    }

	    if(can[k1][k2]){
	    	printf("Data set %d: yes\n",counter++);
	    }else{
	    	printf("Data set %d: no\n",counter++);
	    }
	}

	return 0;
}