POJ2362：Square(DFS)

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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

题意：看给出的所有棍子能不能组成正方形

思路：直接搜索每条边即可

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int a[250],vis[205],sum,l,n;int dfs(int x,int pos,int len){    if(x == 3)        return 1;    int i;    for(i = pos; i>=0; i--)    {        if(!vis[i])        {            vis[i] = 1;            if(len+a[i]<l)            {                if(dfs(x,i-1,len+a[i]))                    return 1;            }            else if(len+a[i]==l)            {                if(dfs(x+1,n-1,0))                    return 1;            }            vis[i] = 0;        }    }    return 0;}int main(){    int t,i,j;    scanf("%d",&t);    while(t--)    {        sum = 0;        scanf("%d",&n);        for(i = 0; i<n; sum+=a[i],i++)            scanf("%d",&a[i]);        l = sum/4;        memset(vis,0,sizeof(vis));        sort(a,a+n);        if(l*4!=sum || n<4 || l<a[n-1])//剪枝        {            printf("no\n");            continue;        }        if(dfs(0,n-1,0))            printf("yes\n");        else            printf("no\n");    }    return 0;}