POJ2362《Square》题解

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

万恶的BZOJ几分钟内就封了，哎，只能先玩POJ了。

分析：dfs可以解决，因为数据量小，所以不用剪枝也能AC。

代码如下：

#include<iostream>#include<cmath>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;int T,N,a[30],f[30];int maxx,minn,sum;bool DFS(int x,int num,int tot){if(tot==sum) num++,tot=x=0;if(minn+tot>sum) return 0;if(num==3) return 1;int last=-1;for(int i=x+1;i<=N;i++)if(!f[i]){if(a[i]==last)continue;f[i]=1;if(DFS(i,num,tot+a[i])) return 1;f[i]=0;last=a[i];}return 0;}int main(){scanf("%d",&T);while(T--){memset(f,0,sizeof(f));scanf("%d",&N);minn=10000000,maxx=sum=0;for(int i=1;i<=N;i++){scanf("%d",&a[i]);sum+=a[i];maxx=max(a[i],maxx);minn=min(a[i],minn);}sort(a+1,a+N+1,greater<int>());if(sum%4){printf("no\n");continue;}sum/=4;if(maxx>sum){printf("no\n");continue;}f[1]=1;if(DFS(1,0,a[1])) printf("yes\n");else printf("no\n");}return 0;}