hdu Problem-4277(dfs+set)

最新推荐文章于 2019-04-26 09:57:29 发布

原创最新推荐文章于 2019-04-26 09:57:29 发布 · 538 阅读

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ACM-STL的使用同时被 2 个专栏收录

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ACM-搜索算法

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探讨如何使用给定长度的围栏段构建不同种类的三角形围栏，确保每种构型的边长各不相同且满足三角形成立条件。采用深度优先搜索算法并利用集合进行重复方案过滤。

USACO ORZ

Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4954 Accepted Submission(s): 1630

Problem Description

Like everyone, cows enjoy variety. Their current fancy is new shapes for pastures. The old rectangular shapes are out of favor; new geometries are the favorite.
I. M. Hei, the lead cow pasture architect, is in charge of creating a triangular pasture surrounded by nice white fence rails. She is supplied with N fence segments and must arrange them into a triangular pasture. Ms. Hei must use all the rails to create three sides of non-zero length. Calculating the number of different kinds of pastures, she can build that enclosed with all fence segments.
Two pastures look different if at least one side of both pastures has different lengths, and each pasture should not be degeneration.

Input

The first line is an integer T(T<=15) indicating the number of test cases.
The first line of each test case contains an integer N. (1 <= N <= 15)
The next line contains N integers li indicating the length of each fence segment. (1 <= li <= 10000)

Output

For each test case, output one integer indicating the number of different pastures.

Sample Input

1
3
2 3 4

Sample Output

//思路：用dfs搜索,然后用set判重(注意代码中的技巧,保证x y z是从小到大排序，且x*1000000+y再放到set中，不能直接x+y放到set中，这样6 6 9和4 8 9就不能正确判断)

AC源码：

#include <iostream>
#include <set>
using namespace std;

typedef long long LL;
LL sum,A[25],n;
set<LL> s;
void dfs(int cnt,int x,int y)
{
	LL z=sum-x-y;
	if(cnt==n)
	{
		if(x<=y&&y<=z&&x+y>z)
		{
	//调试语句	cout<<"x:"<<x<<" y:"<<y<<" z:"<<z<<endl;
			s.insert(x*1000000+y);	
		}
		return ;
	}
	dfs(cnt+1,x+A[cnt],y);
	dfs(cnt+1,x,y+A[cnt]);
	dfs(cnt+1,x,y);
}
int main()
{
	int T;
	cin>>T;
	while(T--)
	{
		s.clear();
		sum=0;
		cin>>n;
		for(int i=0;i<n;++i)
		{
			cin>>A[i];
			sum+=A[i];
		}
		dfs(0,0,0);
		cout<<s.size()<<endl;
	}
	return 0;
}