zoukankan      html  css  js  c++  java
  • POJ 1258 Agri-Net

    题目链接

    Agri-Net

    Time Limit: 1000MS Memory Limit: 10000K
    Total Submissions: 69042 Accepted: 28617
    Description

    Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.
    Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.
    Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.
    The distance between any two farms will not exceed 100,000.

    Input

    The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.

    Output

    For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

    Sample Input

    4
    0 4 9 21
    4 0 8 17
    9 8 0 16
    21 17 16 0

    Sample Output

    28

    思路:最小生成树

    AC代码
    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    using namespace std;
    const int N=110;
    const int inf=0x3f3f3f3f;
    int e[N][N],dis[N],Next[N];
    int n;
    void init(){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(i==j) e[i][j]=0;
                else e[i][j]=inf;
            }
        }
    }
    int Prim(int st){
        int ans=0;
        for(int i=1;i<=n;i++){
            dis[i]=e[st][i];
            Next[i]=st;
        }
        Next[st]=-1;
        int minn,u;
        for(int i=1;i<n;i++){
            minn=inf,u=-1;
            for(int j=1;j<=n;j++){
                if(dis[j]<minn&&Next[j]!=-1){
                    minn=dis[j];
                    u=j;
                }
            }
            if(u==-1 )break;
            ans+=minn;
            Next[u]=-1;
            for(int j=1;j<=n;j++){
                if(dis[j]>e[u][j]&&Next[j]!=-1){
                    dis[j]=e[u][j];
                    Next[j]=u;
                }
            }
        }
        return ans;
    }
    int main(){
        while(~scanf("%d",&n)){
            init();
            for(int i=1;i<=n;i++){
                for(int j=1;j<=n;j++){
                    scanf("%d",&e[i][j]);
                }
            }
            int ans=Prim(1);
            printf("%d
    ",ans);
        }
        return 0;
    }
    
  • 相关阅读:
    eclipse建立工作集管理项目
    echarts-x
    GeoJSON
    mysql 5.7 root password 过期
    kubernetes centos7
    JestClient
    树莓派镜像制作
    docker run elasticsearch
    vm.max_map_count
    远程访问jupyter notebook
  • 原文地址:https://www.cnblogs.com/kun-/p/9714540.html
Copyright © 2011-2022 走看看