"巴卡斯杯" 中国大学生程序设计竞赛 - 女生专场（重现） HDU 5707

Combine String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 17    Accepted Submission(s): 11

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 toa, 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

abcdefadebcfabcdefabecdf

 

Sample Output

YesNo

 

Source

"巴卡斯杯" 中国大学生程序设计竞赛 - 女生专场

 

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题意s1 串 s2串　能够组成s3串。

dp

令f[i][j]代表s1用i了个位置，并且匹配，s2用了j个位置并且匹配。所以每次转移是
 f[i-1][j] &&s1[i]==c[i+j]    f[i][j]=1  or  f[i][j-1]&&s2[j]==c[i+j]

#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;char s1[2005];char s2[2005];char s3[2005];int f[2005][2005];void  work(){    f[0][0]=1;    int i,j;    int l1=strlen(s1+1);    int l2=strlen(s2+1);    int l3=strlen(s3+1);    if(l1+l2!=l3)    {        cout<<"No"<<endl;        return;    }    for(i=0; i<=l1; i++)    {        for(j=0; j<=l2; j++)        {            if(i&&j)                f[i][j]=0;            if(f[i-1][j] &&s3[i+j]==s1[i]&&i)                f[i][j]=1;            if(f[i][j-1]&&s3[i+j]==s2[j]&&j)                f[i][j]=1;        }    }    if(f[l1][l2])        cout<<"Yes"<<endl;    else        cout<<"No"<<endl;}int main(){    while(scanf("%s%s%s",s1+1,s2+1,s3+1)!=EOF)        work();    return 0;}