Square(hdu 1511)

题目描述：

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

 

代码如下：

 /*要剪枝，当所有木棒总长度不能被4整除时以及木棒最大长度大于总长度除以4时，
  * 不能组成正方形，直接输出no
  * 深搜时从第一个开始往后搜索，只要满足当前边长+当前木棒长<正方形边长，
  * 就标记该木棒，并继续搜索后面的木棒，当木棒长度=sum/4 时，count加1，
  * 当count=3时表明能够成正方形，flag=1，返回，flag=0则不能组成正方形。*/
 #include<iostream>
 #include<cstring>

 int visit[];
 bool flag;
 int number[];
 int n,len;

 using namespace std;

 void dfs(int cur,int pos,int count);
 int main()
 {
     int m;
     cin >> m;
     while(m--)
     {
         int max = ,sum = ;
         cin >> n;
         memset(visit,,sizeof(visit));
         for(int i = ;i < n;i++)
         {
             cin >> number[i];
             sum += number[i];
             if(number[i] > max)
                 max = number[i];
         }
         len = sum / ;
         if(sum %  !=  || max > len)//不能总和不能被4整除或最大值大于平均值
         {
             cout << "no" << endl;
             continue;
         }
         flag = ;
         dfs(,,);
         if(flag)
             cout << "yes" << endl;
         else
             cout << "no" << endl;
     }
 }

 void dfs(int cur,int pos,int count)
 {
     if(cur == len)//木棍相加的值等于平均值
     {
         count++;
         cur = ;//当一边成立时，要考虑另外一边注意将cur清0
         pos = ;
         if(count == )//当有三边成立时
         {
             flag = ;
             return;
         }
     }
     for(int i = pos;i < n;i++)
     {
         if(!visit[i])//还没有被访问过
         {
             if((cur + number[i]) <= len)//加上木棍后，长度小于或等于平均值
             {
                 visit[i] = ;
                 dfs(cur + number[i],i,count);
                 if(flag)//一旦有成立的，跳过剩下的判断
                     return;
                 visit[i] = ;//回溯
             }
         }
     }
 }

代码分析：

这道题目要注意的地方就是剪枝。

参考地址：http://www.cnblogs.com/PegasusWang/archive/2013/04/08/3008942.html