Hdu 1501 Zipper【dfs】

Zipper

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9212 Accepted Submission(s): 3263

[align=left]Problem Description[/align]
Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in its original order.

For example, consider forming "tcraete" from "cat" and "tree":

String A: cat

String B: tree

String C: tcraete

As you can see, we can form the third string by alternating characters from the two strings. As a second example, consider forming "catrtee" from "cat" and "tree":

String A: cat

String B: tree

String C: catrtee

Finally, notice that it is impossible to form "cttaree" from "cat" and "tree".

[align=left]Input[/align]
The first line of input contains a single positive integer from 1 through 1000. It represents the number of data sets to follow. The processing for each data set is identical. The data sets appear on the following lines, one data
set per line.

For each data set, the line of input consists of three strings, separated by a single space. All strings are composed of upper and lower case letters only. The length of the third string is always the sum of the lengths of the first two strings. The first two
strings will have lengths between 1 and 200 characters, inclusive.

[align=left]Output[/align]
For each data set, print:

Data set n: yes

if the third string can be formed from the first two, or

Data set n: no

if it cannot. Of course n should be replaced by the data set number. See the sample output below for an example.

[align=left]Sample Input[/align]

3
cat tree tcraete
cat tree catrtee
cat tree cttaree

[align=left]Sample Output[/align]

Data set 1: yes
Data set 2: yes
Data set 3: no

[align=left]Source[/align]
Pacific Northwest 2004 

非常简单的dfs ，结果各种错误都犯了.......

#include<cstdio>
#include<cstring>
const int maxn=1005;
char a[maxn],b[maxn],c[2*maxn];
bool vis[maxn][maxn],kase;
void dfs(int la,int lena,int lb,int lenb)
{
if(la>lena||lb>lenb||vis[la][lb])
{
return;
}
if(la==lena&&lb==lenb)
{
kase=1;
return;
}
vis[la][lb]=1;
if(!kase&&a[la]==c[la+lb])
{
dfs(la+1,lena,lb,lenb);
}
if(!kase&&b[lb]==c[la+lb])
{
dfs(la,lena,lb+1,lenb);
}
}
int main()
{
int t;
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
kase=0;
memset(vis,0,sizeof(vis));
scanf("%s%s%s",a,b,c);
printf("Data set %d: ",i);
dfs(0,strlen(a),0,strlen(b));
printf("%s\n",kase?"yes":"no");
}
return 0;
}

#include<stdio.h>
#include<string.h>
const int maxn=205;
char a[maxn],b[maxn],c[maxn*2];
bool vis[maxn][maxn];
bool ok(int la,int lena,int lb,int lenb)
{
if(la==lena&&lb==lenb)
{
return 1;
}
int kase=0;
if((la<lena||lb<lenb)&&!vis[la][lb])
{
vis[la][lb]=1;
if(a[la]==c[la+lb])
{
kase=ok(la+1,lena,lb,lenb);
}
if(b[lb]==c[la+lb])
{
kase=kase||ok(la,lena,lb+1,lenb);
}
}
return kase;
}
int main()
{
int t,i,j,k,l;
scanf("%d",&t);
for(i=1;i<=t;i++)
{
memset(vis,0,sizeof(vis));
scanf("%s%s%s",a,b,c);
printf("Data set %d: ",i);
printf("%s\n",ok(0,strlen(a),0,strlen(b))?"yes":"no");
}
return 0;
}