计算机学院大学生程序设计竞赛（2015’12）Happy Value（最大生成树）

最高快乐值连接方案

最新推荐文章于 2020-02-04 11:46:33 发布

原创最新推荐文章于 2020-02-04 11:46:33 发布 · 651 阅读

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本文介绍了一种寻找最佳连接方案的问题，旨在通过构建最优的居民间互联网连接来最大化总快乐值。该方案需要考虑每对居民间的友好度，并选择合适的连接路径以确保最高的累积快乐值。

Happy Value

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1350 Accepted Submission(s): 400

Problem Description

In an apartment, there are N residents. The Internet Service Provider (ISP) wants to connect these residents with N – 1 cables.
However, the friendships of the residents are different. There is a “Happy Value” indicating the degrees of a pair of residents. The higher “Happy Value” is, the friendlier a pair of residents is. So the ISP wants to choose a connecting plan to make the highest sum of “Happy Values”.

Input

There are multiple test cases. Please process to end of file.
For each case, the first line contains only one integer N (2<=N<=100), indicating the number of the residents.
Then N lines follow. Each line contains N integers. Each integer H_ij(0<=H_ij<=10000) in i^th row and j^th column indicates that i^th resident have a “Happy Value” H_ijwith j^th resident. And H_ij(i!=j) is equal to H_ji. H_ij(i=j) is always 0.

Output

For each case, please output the answer in one line.

Sample Input

Sample Output

1
8
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <list>
#define LL long long
#define INF 0x3f3f3f3f

using namespace std;

int h[110][110];
int eveco[110];
bool vis[110];
int Pri(int n)
{
    memset(vis,false,sizeof(vis));
    for(int i=1;i<=n;i++)
        eveco[i] = h[1][i];
    int cost = 0;
    vis[1] = true;
    eveco[1] = 0;
    for(int i=1;i<n;i++)
    {
        int m = 0,pos;
        for(int j=1;j<=n;j++)
        {
            if(m < eveco[j] && !vis[j])
                pos = j, m= eveco[j];
        }
        cost += m;
        vis[pos] = true;
        for(int j=1;j<=n;j++)
            if(!vis[j] && eveco[j] < h[pos][j])
            eveco[j] = h[pos][j];
    }
    return cost;
}


int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                cin>>h[i][j];
            }
        }
        cout<<Pri(n)<<endl;
    }
    return 0;
}