hdu 4277 USACO ORZ dfs+hash

最新推荐文章于 2021-10-21 18:26:48 发布

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最新推荐文章于 2021-10-21 18:26:48 发布

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文章标签： java

原文链接：http://www.cnblogs.com/jhz033/p/7575553.html

本文介绍了一道USACOORZ竞赛题目，任务是利用给定长度的围栏段构建不同类型的三角形牧场。文章提供了完整的源代码实现，并详细解释了如何通过递归深度优先搜索算法来解决此问题。

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USACO ORZ

Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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

Source

2012 ACM/ICPC Asia Regional Changchun Online

题意：用所有点组成三角形，问不同的个数；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<bitset>
#include<set>
#include<map>
#include<time.h>
using namespace std;
#define LL long long
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e5+10,M=1e6+10,inf=1e9+10;
const LL INF=1e18+10,mod=1e9+7;
const double eps=(1e-8),pi=(4*atan(1.0));

int num[N],n;
struct hashnum
{
    const static int si=1e6+7;
    vector<LL>v[1000010];
    void add(LL x)
    {
        LL temp=x;
        x%=si;
        for(int i=0;i<v[x].size();i++)
            if(v[x][i]==temp)return;
        v[x].push_back(temp);
    }
}hs;
void dfs(int pos,int a,int b,int c)
{
    if(pos>n)
    {
        if(a>b)swap(a,b);
        if(b>c)swap(b,c);
        if(a>b)swap(a,b);
        if(a+b>c)
        {
            hs.add(1LL*a*10000000+b);
        }
        return;
    }
    dfs(pos+1,a+num[pos],b,c);
    dfs(pos+1,a,b+num[pos],c);
    dfs(pos+1,a,b,c+num[pos]);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        for(int i=0;i<=hs.si;i++)
            hs.v[i].clear();
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);
        dfs(1,0,0,0);
        int ans=0;
        for(int i=0;i<hs.si;i++)
            ans+=hs.v[i].size();
        printf("%d\n",ans);
    }
    return 0;
}