HDU 1518 DFS
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Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11432 Accepted Submission(s): 3671
Problem Description
Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?
Input
The first line of input contains N, the number of test cases. Each test case begins with an integer 4 <= M <= 20, the number of sticks. M integers follow; each gives the length of a stick - an integer between 1 and 10,000.
Output
For each case, output a line containing "yes" if is is possible to form a square; otherwise output "no".
Sample Input
34 1 1 1 15 10 20 30 40 508 1 7 2 6 4 4 3 5
Sample Output
yesnoyes
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int s[10003]={0};
bool mark[10003]={0};
int m=0;
bool dfs(int len,int l,int count,int pos);
bool cmp(int x,int y)
{
return x>y;
}
int main()
{
int T=0;
scanf("%d",&T);
while(T--)
{
int sum=0;
scanf("%d",&m);
for(int i=0;i<m;i++)
{
scanf("%d",&s[i]);
sum+=s[i];
mark[i]=0;
}
if(sum%4||m<4)
{
printf("no\n");
continue;
}
sum/=4;
sort(s,s+m,cmp);
if(dfs(sum,0,0,0))
{
printf("yes\n");
}
else
{
printf("no\n");
}
}
return 0;
}
bool dfs(int len,int l,int count,int pos)
{
if(count==4) return 1;
for(int i=pos;i<m;i++)
{
if(mark[i]) continue;
if(l+s[i]==len)
{
mark[i]=1;
if(dfs(len,0,count+1,0))
return 1;
mark[i]=0;
}
else if(l+s[i]<len)
{
mark[i]=1;
if(dfs(len,l+s[i],count,i+1))
return 1;
mark[i]=0;
}
}
return 0;
}
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int s[10003]={0};
bool mark[10003]={0};
int m=0;
bool dfs(int len,int l,int count,int pos);
bool cmp(int x,int y)
{
return x>y;
}
int main()
{
int T=0;
scanf("%d",&T);
while(T--)
{
int sum=0;
scanf("%d",&m);
for(int i=0;i<m;i++)
{
scanf("%d",&s[i]);
sum+=s[i];
mark[i]=0;
}
if(sum%4||m<4)
{
printf("no\n");
continue;
}
sum/=4;
sort(s,s+m,cmp);
if(dfs(sum,0,0,0))
{
printf("yes\n");
}
else
{
printf("no\n");
}
}
return 0;
}
bool dfs(int len,int l,int count,int pos)
{
if(count==4) return 1;
for(int i=pos;i<m;i++)
{
if(mark[i]) continue;
if(l+s[i]==len)
{
mark[i]=1;
if(dfs(len,0,count+1,0))
return 1;
mark[i]=0;
}
else if(l+s[i]<len)
{
mark[i]=1;
if(dfs(len,l+s[i],count,i+1))
return 1;
mark[i]=0;
}
}
return 0;
}
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