HDU 1518--Square(DFS)

 Square

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12479 Accepted Submission(s): 3971

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

剪枝条件为判断符合条件的边数是否为4个，其次当一个边满足条件，需要从头重新搜索。。

以下AC代码：

#include<stdio.h>#include<string.h>int n,m;int a[25],flag[25];int mid;int temp;void dfs(int sum,int index,int count){    if (count==4)    {        temp=1;        return;    }    if(sum==mid)    {        dfs(0,0,count+1);        if(temp)        return;    }    int i;    for(i=index;i<m;i++)    {        if(!flag[i] && sum+a[i]<=mid)        {            flag[i]=1;            dfs(sum+a[i],i+1,count);            if(temp)                return;            flag[i]=0;        }    }}int main(){    int i;    int sum;    scanf("%d",&n);    while(n--)    {        sum=0;        scanf("%d",&m);        for(i=0;i<m;i++)        {            scanf("%d",&a[i]);            sum+=a[i];        }        if(sum%4)        {            printf("no\n");            continue;        }        else        {            memset(flag,0,sizeof(flag));            temp=0;            mid=sum/4;            dfs(0,0,0);        }            if(temp)               printf("yes\n");            else               printf("no\n");    }    return 0;}