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最新推荐文章于 2024-06-19 14:52:40 发布

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本文探讨了使用不同长度的棍子能否拼接成正方形的问题。通过递归深度优先搜索算法，结合剪枝策略，有效地解决了该问题。文章提供了一个完整的C++实现案例。

Square
Time Limit : 10000/5000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 31 Accepted Submission(s) : 7
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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
//减掉最大值超过平均值，和sum无法整除的情况，然后每次当前值=aver时count+1就行了，当count==4时结束，否则dfs。

#include<iostream>
#include<cstdio>
#define max(a,b) (a)>(b)?(a):(b)
using namespace std;

int str[23],n,sum,aver,maxn;
bool vist[23],flag;
void dfs(int mySum,int Count,int t)
{
    if(mySum==aver)
    {
       Count++;
       if(Count==4)
       {
          flag=true;
          return;
       }
       else mySum=0;
       t=0;

    }
    if(flag) return;
    for(int i=t;i<n;i++)
    {
       if(!vist[i]&&mySum+str[i]<=aver)
       {
          vist[i]=true;
          dfs(mySum+str[i],Count,i);
          vist[i]=false;
       }
    }

}
int main()
{
   int T,i;
   scanf("%d",&T);
   while(T--)
   {
      scanf("%d",&n);
      sum=maxn=0;
      for(i=0;i<n;i++)
      {
         scanf("%d",&str[i]);
         sum+=str[i];
         maxn=max(str[i],maxn);
         vist[i]=false;
      }
      if(sum%4!=0||maxn>sum/4)//简单判断、不满足的直接给出结果
          printf("no\n");
      else
      {
         flag=false;
         aver=sum/4;
         dfs(0,0,0);
         if(flag) printf("yes\n");
         else printf("no\n");
      }
   }
   return 0;
}