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_ij with j^thresident. 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

0 1

1 0

0 1 5

1 0 3

5 3 0

Sample Output

最大生成树

把最下生成树的算法倒着来就行。

比如把权值改成负数，结果再换回来

#include<stdio.h>

//#include<bits/stdc++.h>

#include<string.h>

#include<iostream>

#include<math.h>

#include<sstream>

#include<set>

#include<queue>

//#include<map>

#include<vector>

#include<algorithm>

#include<limits.h>

#define inf 0x3fffffff

#define INF 0x3f3f3f3f

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define ULL unsigned long long

using namespace std;

int map[110][110];

int dis[110];

int flag[110];

int n;

int Prime()

{

    int i,j;

    int pos,sum=0;

    memset(flag,0,sizeof(flag));

    for(i=1; i<=n; i++)

    {

        dis[i]=map[1][i];

    }

    flag[1]=1;

    for(i=1; i<n; i++)

    {

        int mid=inf;

        for(j=1; j<=n; j++)

        {

            if(!flag[j]&&dis[j]<mid)

            {

                mid=dis[j];

                pos=j;

            }

        }

        flag[pos]=1;

        sum+=mid;

        for(j=1; j<=n; j++)

        {

            if(!flag[j]&&map[pos][j]<dis[j])

            {

                dis[j]=map[pos][j];

            }

        }

    }

    return sum;

}

int main()

{

    int i,j;

    int q,a,b;

    while(~scanf("%d",&n))

    {

        memset(map,0,sizeof(map));

        int ans;

        for(i=1; i<=n; i++)

        {

            for(j=1; j<=n; j++)

            {

                scanf("%d",&ans);

                map[i][j]=-ans;

            }

        }

        printf("%d\n",-Prime());

    }

    return 0;

}

巴特西

HDU计算机学院大学生程序设计竞赛（2015’12）Happy Value

Problem Description

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