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
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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;
}
