zoukankan      html  css  js  c++  java
  • 最小生成树 D

    There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected. 

    We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

    Input

    The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j. 

    Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

    Output

    You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

    Sample Input

    3
    0 990 692
    990 0 179
    692 179 0
    1
    1 2
    

    Sample Output

    179
    #include<iostream>
    #include<string>
    #include<algorithm>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<iomanip>
    using namespace std;
    #define MAXN 104
    #define INF 0x3f3f3f3f
    bool been[MAXN];
    int n,q,g[MAXN][MAXN],lowcost[MAXN];
    /*
    最小生成树,如果D(a,b)<=ra+rb,那么g[a][b]=0
    否则D(a,b)>ra+rb,g[a][b] = ra+rb-D(a,b)
    */
    int Prim(int beg)
    {
        int ans = 0;
        memset(been,false,sizeof(been));
        for(int i=1;i<=n;i++)
        {
            if(i==beg) lowcost[i] = 0;
            else lowcost[i] = g[beg][i];
        }
        been[beg] = true;
        for(int j=1;j<n;j++)
        {
            int Minc = INF,k=-1;
            for(int i=1;i<=n;i++)
            {
                if(!been[i]&&lowcost[i]<Minc)
                {
                    Minc = lowcost[i];
                    k = i;
                }
            }
            if(k==-1) return -1;
            been[k] = true;
            ans+=Minc;
            for(int i=1;i<=n;i++)
            {
                if(!been[i]&&lowcost[i]>g[k][i])
                    lowcost[i] = g[k][i];
            }
        }
        return ans;
    }
    int main()
    {
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                cin>>g[i][j];
            }
        }
        cin>>q;
        int tx,ty;
        for(int i=1;i<=q;i++)
        {
            cin>>tx>>ty;
            g[tx][ty] = g[ty][tx] = 0;
        }
        int ans = Prim(1);
        cout<<ans<<endl;
        return 0;
    }
  • 相关阅读:
    Linux 内核中的 Device Mapper 机制
    阿里云 Angular 2 UI框架 NG-ZORRO介绍
    Docker容器 暴露多个端口
    修改docker容器的端口映射
    Ubuntu Docker安装
    Docker容器技术的PaaS云平台架构设计***
    scala 学习笔记三 闭包
    scala 学习笔记二 方法与函数
    scala 学习笔记一 列表List
    Python3 写Windows Service服务程序
  • 原文地址:https://www.cnblogs.com/joeylee97/p/6589896.html
Copyright © 2011-2022 走看看