Hdu 1501 Zipper

Zipper

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

Total Submission(s): 4209    Accepted Submission(s): 1513


[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
 

[align=left]Recommend[/align]
linle

#include<stdio.h>
char a[201],b[201],c[402];
int visit[201][201]={0};			//剪枝
int  isOk=0;
void dfs(int i,int j,int k)
{	if(visit[i][j]==1) return;			//剪枝
if(c[k]=='\0')
{	isOk=1;	return;	}
if(a[i]!='\0'&&c[k]==a[i])
dfs(i+1,j,k+1);
if(b[j]!='\0'&&c[k]==b[j])
dfs(i,j+1,k+1);
visit[i][j]=1;				//剪枝
}
int main()
{	int n,i,j,cnt=0;
//	int flag=0;
//	freopen("d:\\data.in.txt","r",stdin);
scanf("%d",&n);
while(n--)
{	isOk=0;
for(i=0;i<201;i++)            //剪枝
for(j=0;j<201;j++)		//剪枝
visit[i][j]=0;		//剪枝
scanf("%s%s%s",a,b,c);
dfs(0,0,0);
printf("Data set %d: %s\n",++cnt,isOk?"yes":"no");
}
return 0;
}

感觉dp应该更快，就写了个dp的程序，结果比dfs还慢，应该是程序样例给的太简单，没有出复杂的数据。

#include<stdio.h>
#include<string.h>

#define MAXN 204

char a[MAXN],b[MAXN],c[2*MAXN];
int d[MAXN][MAXN];

int main(void)
{
int T,i,j,t;
int d1,d2,d3;
t=1;
//	freopen("d:\\in.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
scanf("%s%s%s",a,b,c);
d1=strlen(a);
d2=strlen(b);
for(i=0;i<=d1;i++)
for(j=0;j<=d2;j++)
d[i][j]=0;
d[0][0]=1;
for(i=1;i<=d1;i++)
{
if(a[i-1]==c[i-1])
d[i][0]=1;
else
break;
}
for(i=1;i<=d2;i++)
{
if(b[i-1]==c[i-1])
d[0][i]=1;
else
break;
}
for(i=1;i<=d1;i++)
for(j=1;j<=d2;j++)
{
if(d[i][j-1] && b[j-1]==c[i+j-1])
d[i][j]=1;
if(d[i-1][j] && a[i-1]==c[i+j-1])
d[i][j]=1;
}
printf("Data set %d: ",t++);
if(d[d1][d2])
printf("yes\n");
else
printf("no\n");
}
return 0;
}