hdu-1518-Square-深搜+剪枝（同hdu1455）

Square

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

给出m个数代表木棍的长度，问这m条木棍能否形成一个正方形

1、先将m个数从大到小排列

2、将m个数之和sum，x=sum/4作为边长，进行搜索，

三个参数:(1)形成一条边长时的当前长度su，(初始为0) (2)形成边的数目cc (3)下一次搜索的位置pos

对于(3),当su+s[i]<x,下一次应从i+1开始搜，因为前面满足的都已经标记过了,而当su+s[i]==x时说明已经找到一条边了,则下一次应重新从0开始找

对于su+s[i]==x,当s[i]不满足,则后面所有能形成s[i]的都不满足

对于su==0,说明s[i]作为第一条木棍不能满足，则在其他边也不能满足

s[i]==s[i+1] ,如果s[i]不满足，则与它相等的s[i+1]也不用再搜了

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int s[50],sum,m,x;
int vis[50];
int cmp(int aa,int bb)
{
return aa>bb;
}
int dfs(int su,int cc,int pos)
{
if(cc==4) return 1;
if(su == x)
{
su=0;
cc++;
}
for(int i=pos ; i < m ; i++)
{
if(vis[i] == 0 && su+s[i]<=x)
{
vis[i]=1;
if(su+s[i]<x)
{
if(dfs(su+s[i],cc,i+1)) return 1;
}
else if(su+s[i]==x)
{
if(dfs(0,cc+1,0)) return 1;
}
vis[i]=0;
if(su+s[i]==x || su==0 ) break;
while(i+1<m && s[i]==s[i+1])
{
i++;
}
}
}
return 0;
}
int main()
{
int n,T,i;
scanf("%d",&T);
while(T--)
{
memset(vis,0,sizeof(vis));
scanf("%d",&m);
sum = 0;
for(i=0 ; i < m ; i++)
{
scanf("%d",&s[i]);
sum+=s[i];
}
sort(s,s+m,cmp);
x=sum/4;
if(sum%4==0 && dfs(0,0,0))
printf("yes\n");
else
printf("no\n");
}
return 0;
}