POJ2192[Zipper]

动态规划，f[i][j]表示用a串的前i个字符和b串的前j个字符匹配c串的前i+j个字符是否可行。

那么f[i][j]=(f[i][j-1] and b[j]==c[i+j]) or (f[i-1][j] and a[i]==c[i+j])，边界f[0][0]=true。

CODE:

/*

AUTHOR: Su Jiao

DATE: 2010-7-22

DESCRIPTION:

动态规划 乱做

http://acm.pku.edu.cn/JudgeOnline/problem?id=2192

*/

#include <stdio.h>

#include <string.h>

 

const int MAXLENGTH=400+1;

typedef char string[MAXLENGTH];

int n;

string a,b,c;

int f[MAXLENGTH][MAXLENGTH];

string s[]={“no”,“yes”};

 

int main()

{

    scanf(“%d”,&n);

    for (int nn=1;nn<=n;nn++)

    {

        scanf(“%s%s%s”,&a,&b,&c);

        int al=strlen(a),bl=strlen(b),cl=strlen(c);

        for (int i=0;i<=al;i++)

            for (int j=0;j<=bl;j++)

            {

                f[i][j]=false;

                if (j>0&&b[j-1]==c[i+j-1]&&f[i][j-1]) f[i][j]=true;

                if (i>0&&a[i-1]==c[i+j-1]&&f[i-1][j]) f[i][j]=true;

                if (!i&&!j) f[i][j]=true;

            }

        printf(“Data set %d: %s\n”,nn,s[al+bl==cl&&f[al][bl]]);

    }

}