Poj 2192 zipper

Today, I suddenly found out that I was not very impressed with this question, but I thought it was a classic question. So I decided to write a conclusion report.

The first question is to give you three strings. The first two strings can be constructed.

But you need to know that this condition is

They cannot change the order of the two final strings. </P> <p> two state equations are obtained.

DP [I] [J] indicates whether the first I character and the first J character constitute the first I + J characters of S3

(1)dp[i][j]=(dp[i-1][j]&&s1[i]==s3[i+j])?ture:flase

(2)dp[i][j]=(dp[i][j-1]&&s2[j==s3[i+j])?true:flase

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;char s1[500],s2[500],s3[500];int dp[500][500];int main(){    int t,i,j,k,len1,len2,len3,cas = 1 ;    scanf("%d",&t);    while(t--)    {        scanf("%s%s%s",s1+1,s2+1,s3+1);        len1 = strlen(s1+1);        len2 = strlen(s2+1);        len3 = strlen(s3+1);        memset(dp,0,sizeof(dp));        for(i = 1; i<=len1; i++)        {            if(s1[i] == s3[i])                dp[i][0] = 1;            else break;        }        for(i = 1; i<=len2; i++)        {            if(s2[i] == s3[i])                dp[0][i] = 1;            else break;        }        for(i = 1; i<=len1; i++)        {            for(j = 1; j<=len2; j++)            {                if(s3[i+j] == s1[i] && dp[i-1][j])                    dp[i][j] = 1;                if(s3[i+j] == s2[j] && dp[i][j-1])                    dp[i][j] = 1;            }        }        printf("Data set %d: ",cas++);        if(dp[len1][len2])            printf("yes\n");        else            printf("no\n");    }    return 0;}

Poj 2192 zipper