hdu 5707 Combine String

Combine String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2945    Accepted Submission(s): 823

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

 

Source

"巴卡斯杯" 中国大学生程序设计竞赛 - 女生专场 

 

Recommend

liuyiding

 

Statistic |  Submit |  Discuss |  Note

这题的意思是s1和s2字符串能否拼接成s3字符串，s1和s2内的字符相对位子不能打乱

这题刚开始做的时候感觉是动态规划，但是又推不出动态规划的转化公式，后来看了别人写的题解，发现这题很巧妙

其实这题就像是最长公共子序列（LCS）

转化方程为

s1[i]==s3[I+j].    则dp[i][j]=max(dp[i][j],dp[i-1][j])

s2[j]==s3[I+j].   则dp[i][j]=max(dp[i][j],dp[i][j-1])

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char s1[2005],s2[2005],s3[2005];
int v[2005][2005];
int main() {
    while (scanf("%s",s1+1)!=EOF) {
        scanf("%s%s",s2+1,s3+1);
        memset(v, 0, sizeof(v));
        v[0][0]=1;
        int t1=int(strlen(s1+1)),t2=int(strlen(s2+1));
        if((t1+t2)!=strlen(s3+1)){
            printf("No\n");
            continue;
        }
        for(int i=0;i<=t1;i++)
            for (int j=0; j<=t2; j++) {
                if(i>0&&s1[i]==s3[i+j])
                    v[i][j]=max(v[i-1][j],v[i][j]);
                if(j>0&&s2[j]==s3[i+j])
                    v[i][j]=max(v[i][j-1],v[i][j]);
            }
        if(v[t1][t2])
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}