Square
Time Limit : 10000/5000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 20 Accepted Submission(s) : 12
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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
3
4 1 1 1 1
5 10 20 30 40 50
8 1 7 2 6 4 4 3 5
Sample Output
yes
no
yes
Source
University of Waterloo Local Contest 2002.09.21
此题需要优化时间，避免超时。。优化时间技巧可以学习。。。。。。。。。
#include <stdio.h>
#include<string.h>
int a[];
int vist[];
int sum;
int l;
int n;
int flag;
void Dfs(int t, int len, int index)
{ if (t == )
{
flag = ;
return ;
} if (len == l)
{
Dfs(t + , , );
if (flag)//优化时间
{
return ;
}
} for (int i = index; i < n; i++)//从index开始优化时间
{
if (vist[i]== && a[i] + len <= l)
{
vist[i] = ;
Dfs(t, a[i] + len, i + );
if (flag)//优化时间
{
return;
}
vist[i] = ;
}
}
} int main()
{
int i,t;
scanf("%d", &t);
while (t--)
{ sum = ;
scanf("%d", &n);
for (int i = ; i < n; i++)
{
scanf("%d", &a[i]);
sum += a[i];
} if (sum % != )//简答的优化
{
puts("no");
continue;
} l = sum / ; for (i = ; i < n; i++)//有比边长大的边就不行
{
if (a[i] > l)
{
break;
}
}
if (i != n)
{
puts("no");
continue;
}
memset(vist, , sizeof(vist));
flag = ;
Dfs(, , );
if (flag)
{
puts("yes");
}
else
{
puts("no");
}
}
return ;
}