POJ 2362(Square-搜索剪枝1-相对顺序)
Language:DefaultSquareTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 17066 Accepted: 5878

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

Source

Waterloo local 2002.09.21

《搜索是怎样剪枝的-1》

1.只要找到3条边。

2.从大到小顺序找。

3.每次搜边时要确定边的相对顺序唯一（直接从TLE→秒）

#include<cstdio>#include<cstdlib>#include<algorithm>#include<functional>#include<cstring>#include<iostream>using namespace std;#define MAXM (20+10)int tt,n,a[MAXM],cnt,len;bool b[MAXM],flag;bool dfs(int l,int x,int pre){//cout<<l<<' '<<x<<' '<<kth<<endl;if (x==len) {l++;x=0;pre=n-1;}if (l==4){return 1;}for(int i=pre-1;i;i--)if (!b[i]&&x+a[i]<=len){b[i]=1;if (dfs(l,x+a[i],i)) return 1;b[i]=0;//if (!x) return 0;}return 0;}int main(){scanf("%d",&tt);while (tt--){cnt=0;scanf("%d",&n);for (int i=1;i<=n;i++){scanf("%d",&a[i]);cnt+=a[i];b[i]=0;}sort(a+1,a+1+n);if (n<4||cnt%4||a[n]>cnt/4){printf("no\n");continue;}b[n]=1;len=cnt/4;if (dfs(1,a[n],n)) {printf("yes\n");}else printf("no\n");}return 0;}