hdu5707

Combine String

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 27    Accepted Submission(s): 17
[/b]

[align=left]Problem Description[/align]
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''.

 

[align=left]Input[/align]
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).

 

[align=left]Output[/align]
For each test case, print ``Yes'', if
c
is a combine string of a
and b,
otherwise print ``No''.

 

[align=left]Sample Input[/align]

abc
def
adebcf
abc
def
abecdf

 

[align=left]Sample Output[/align]

Yes
No

本题的DP类似于LCS（longest common subsequence）（如果不了解的可以搜一下）。

只是，

f[i][j]表示第一个字符串用了前i个位置（第i个位置已匹配），第二个字符串的前j个位置（第j个位置已匹配）

是否可以对c串成功匹配（成功匹配则必然会匹配到c串的前i+j个位置）。

f[i][j]==1则表示可以成功匹配

f[i][j]==0则表示无法成功匹配

显然，初始只有f[0][0]==1

所以，我们有——

f[i][j]= f[i-1][j]&&(a[i]==c[i+j])

|| f[i][j-1]&&(b[j]==c[i+j])

这样，最终f
[m]==1则为Yes否则为No

【时间复杂度&&优化】

O(nm)

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int N = 2005;
char a
,b
,c
;
int dp

;
int main()
{
int i,j,n,m,k;
while(~scanf("%s",a+1))
{
memset(dp,0,sizeof(dp));
scanf("%s",b+1);
scanf("%s",c+1);
n=strlen(a+1);
m=strlen(b+1);
k=strlen(c+1);
if(n+m!=k)
{
puts("No");
continue;
}
dp[0][0]=1;
for(i=0;i<=n;i++)
for(j=0;j<=m;j++)
{
if(i>0&&c[i+j]==a[i])
dp[i][j]|=dp[i-1][j];
if(j>0&&c[i+j]==b[j])
dp[i][j]|=dp[i][j-1];
}
if(dp
[m])
puts("Yes");
else
puts("No");
}
return 0;
}