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
题意如下:
询问能否用这些木棍拼成正方形。
这个题需要剪枝。
剪枝的情况如下:
(1)总的和余4不等于0,那么一定不能构成
(2)最大值大于他们的平均值,那么一定不能构成
(3)如果三个边都等于平均值,那么一定能构成三角形。
需要注意的是,如果找出一条边的话,那么直接从头开始找。
反之,继续遍历。
代码如下:
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std;
int m;
int a[25];
int is[25];
int flag=0;
int ave;
int sum;
void dfs (int loc,int ssum,int num)
{
if(num==3)
{
printf("yes\n");
flag=1;
return;
}
if(ssum==ave)
{
dfs(0,0,num+1);
if(flag)
return;
}
for (int i=loc;i<m;i++)
{
if(!is[i]&&ssum+a[i]<=ave)
{
is[i]=1;
dfs(i,ssum+a[i],num);
is[i]=0;
}
if(flag)
return;
}
}
int main()
{
int n;
scanf("%d",&n);
while (n--)
{
flag=0;
memset (is,0,sizeof(is));
sum=0;
scanf("%d",&m);
int Max=0;
for (int i=0;i<m;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
Max=max(a[i],Max);
}
if(sum%4||Max>sum/4)
printf("no\n");
else
{
ave=sum/4;
dfs(0,0,0);
if(!flag)
printf("no\n");
}
}
return 0;
}