【POJ】2362 Square(DFS)

【题目链接】http://poj.org/problem?id=2362

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题目内容

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 asquare; 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

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题目大意

给出n根木棍问能不能拼出正方形

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解题思路

如果木棍长度总和不是4的倍数，直接排除，不然长度和/=4，如果能dfs构成三条边，就能构成正方形。
具体的dfs过程见注释。

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AC代码

    #include<iostream>    #include<cstdio>    #include<cstring>    #include<cstdlib>    #include<string>    #include<algorithm>    #include<math.h>    #include<limits.h>    #include<stack>    #include<queue>    #define LL long long    using namespace std;    int a[25];    int b[25];标记木棍i有没有被用过    bool cmp(int a,int b)    {        return a>b;    }    int n,sum;    bool flag;    void dfs(int x,int i,int num)//x是已经拼好的长度，i是待检测的木棍，num是已经拼好的边的数量；    {        if(num==3||flag)            {                flag=true;                return;            }            for(int j=i;j<n;j++)            {                if(a[j]>sum-x) return;                if(b[j]==0&&a[j]==sum-x)                {                    b[j]=1;                    dfs(0,0,num+1);                    b[j]=0;                    return;                }                if(b[j]==0&&a[j]<sum-x)                {                    b[j]=1;                    dfs(x+a[j],j+1,num);                    if(flag) return;                    b[j]=0;                }            }    }    int main()    {        int t;        scanf("%d",&t);        while(t--)        {            sum=0;            flag=false;            memset(b,0,sizeof(b));            scanf("%d",&n);            for(int i=0;i<n;i++)            {                scanf("%d",&a[i]);                sum+=a[i];            }            if(sum%4!=0)            {                printf("no\n");                continue;            }//排除总长度不是4的倍数的情况            sum/=4;            sort(a,a+n);            if(a[n-1]>sum)            {                printf("no\n");                continue;            }            dfs(0,0,0);            if(flag)                printf("yes\n");            else printf("no\n");        }        return 0;    }