Problem Description
Given three strings , and , your mission is to check whether is the combine string of and .
A string is said to be the combine string of and if and only if can be broken into two subsequences, when you read them as a string, one equals to , and the other equals to .
For example, ``adebcf'' is a combine string of ``abc'' and ``def''.
A string is said to be the combine string of and if and only if can be broken into two subsequences, when you read them as a string, one equals to , and the other equals to .
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 , and (the length of each string is between 1 and 2000).
Each test case contains three strings , and (the length of each string is between 1 and 2000).
Output
For each test case, print ``Yes'', if is a combine string of and , otherwise print ``No''.
Sample Input
abc
def
adebcf
abc
def
abecdf
直接用递归来暴力,但是如果有的字符是相同的,就出现多种情况,走着走着可能就要回溯到某一步,这就很浪费时间,所以应该采取用空间换时间的动态规划算法。我们可以建立一个二维数组G,G的行 i 列 j 数是代表a的前i个成功匹配到了c,b的前j个成功匹配到了c,合起来就是a和b成功匹配了i+j-1个c字符,最后只要检验这个二维数组G【len_a】【len_b】这个点是否等于1,就知道能不能匹配到了。代码:
#include<stdio.h>
#include<string.h>
char a[2010],b[2010],c[2010];
int maze[2010][2010];
int main()
{
int i,j,la,lb,lc;
while(scanf("%s %s %s",a,b,c)!=EOF){
la=strlen(a);
lb=strlen(b);
lc=strlen(c);
if(lc!=(la+lb)){
printf("No\n");
continue;
}
memset(maze,0,sizeof(maze));
maze[0][0]=1;
for(i=1;i<=la;i++){
if(a[i-1]==c[i-=1]&&maze[0][i-1]==1)
maze[0][i]=1;
else
break;
}
for(i=1;i<=lb;i++){
if(b[i-1]==c[i-1]&&maze[i-1][0]==1)
maze[i-1][0]=1;
else
break;
}
for(i=1;i<=lb;i++){
for(j==1;j<=la;j++){
if((maze[i-1][j]==1&&a[j-1]==c[i+j-1])||(maze[i][j-1]==1&&b[i-1]==c[i+j-1]))
maze[i][j]=1;
else
break;
}
}
if(maze[la][lb]==1)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}
Sample Output