zoukankan      html  css  js  c++  java
  • 最小生成树模板(poj3625)

    Building Roads
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 9360   Accepted: 2690

    Description

    Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms.

    Each of the N (1 ≤ N ≤ 1,000) farms (conveniently numbered 1..N) is represented by a position (XiYi) on the plane (0 ≤ X≤ 1,000,000; 0 ≤ Y≤ 1,000,000). Given the preexisting M roads (1 ≤ M ≤ 1,000) as pairs of connected farms, help Farmer John determine the smallest length of additional roads he must build to connect all his farms.

    Input

    * Line 1: Two space-separated integers: N and M
    * Lines 2..N+1: Two space-separated integers: Xand Y
    * Lines N+2..N+M+2: Two space-separated integers: i and j, indicating that there is already a road connecting the farm i and farm j.

    Output

    * Line 1: Smallest length of additional roads required to connect all farms, printed without rounding to two decimal places. Be sure to calculate distances as 64-bit floating point numbers.

    Sample Input

    4 1
    1 1
    3 1
    2 3
    4 3
    1 4

    Sample Output

    4.00

    Source



    prim算法:

    Memory: 8072K   Time: 188MS
    Language: C++   Result: Accepted
    #include <iostream>
    #include <cmath>
    #include <cstring>
    #include <cstdio>
    #define INF 0x3f3f3f3f
    using namespace std;
    const int N = 1001;
    double graph[N][N];
    bool visit[N];
    int n,M;
    typedef struct
    {
        double x;
        double y;
    }dian;
    dian m[N];
    
    double prim()
    {
    
        memset(visit,0,sizeof(visit));
    
        double low[1001];
        int pos = 1;
        visit[1] = 1;
        double result = 0;
    
        for(int i = 2; i <= n; i++)
        {
            low[i] = graph[pos][i];
        }
    
        for(int i = 0; i < n-1; i++)
        {
            double Min = INF;
    
            for(int j = 1; j <= n; j++)
            {
                if(!visit[j] && Min > low[j])
                {
                    Min = low[j];
                    pos = j;
    
                }
            }
            visit[pos] = 1;
            result += Min;
    
            for(int i = 1; i <= n; i++)
            {
                if(!visit[i] && low[i] > graph[pos][i])
                {
                    low[i] = graph[pos][i];
                }
            }
    
        }
        return result;
    }
    
    double dis(dian a,dian b)
    {
        return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
    }
    
    int main()
    {
       // freopen("in.txt","r",stdin);
        while(cin>>n>>M)
        {
            memset(graph,INF,sizeof(graph));
            for(int i = 1; i <=n;i++)
            {
                cin >> m[i].x>>m[i].y;
            }
            for(int i = 1; i <= n; i++)
            {
                for(int j = i + 1; j <= n; j++)
                {
                    graph[i][j] = graph[j][i] = dis(m[i],m[j]);
                }
            }
            for(int i = 0; i < M ; i++)
            {
                int a,b;
                cin>>a>>b;
                graph[a][b] = graph[b][a] = 0;
            }
            printf("%.2lf
    ",prim());
        }
        return 0;
    }


    kruskal算法

    Memory: 8604K   Time: 735MS
    Language: G++   Result: Accepted
    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    const int N = 1001;
    const int E = 1000000;
    int n, M;
    int cent;
    int a[N];
    int Count = 0;
     
    typedef struct
    {
        int x;
        int y;
        double vaule;
    }dian;
    dian m[E];
     
    typedef struct
    {
        double x, y;
    }situation;
    situation p[N];
     
    double dis(situation a, situation b)
    {
        return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y) * (a.y - b.y));
    }
     
    bool cmp(dian a, dian b)
    {
        return a.vaule < b.vaule;
    }
     
    void init()
    {
        //                   cent 这里应该初始化到n
        for (int i = 1; i <= n; i++)
        {
            a[i] = i;
        }
    }
     
    int Find(int x)
    {
        while (x != a[x])
        {
            x = a[x];
        }
        return x;
    }
     
    void Union(int x, int y)
    {
        // 建议做路径压缩
        int fx = Find(x);
        int fy = Find(y);
        if (fx != fy)
        {
            a[fx] = fy;
        }
    }
     
    double Kruskal()
    {
        // init(); 不应该在这里init
        sort(m, m + cent, cmp);
        double result = 0;
        for (int i = 0; i < cent&&Count != n - 1; i++)
        {
            if (Find(m[i].x) != Find(m[i].y))
            {
                Union(m[i].x, m[i].y);
                result += m[i].vaule;
                Count++;
            }
        }
        return result;
    }
     
    int main()
    {
        while (cin >> n >> M)
        {
           
            for (int i = 1; i <= n; i++)
            {
                cin >> p[i].x >> p[i].y;
            }
            cent = 0;
            Count = 0;
            for (int i = 1; i <= n; i++)
            {
                for (int j = i + 1; j <= n; j++)
                {
                    m[cent].x = i;
                    m[cent].y = j;
                    m[cent++].vaule = dis(p[i], p[j]);
                }
            }
            // init不应该放在Kruskal里面
            init();
            for (int i = 1; i <= M; i++)
            {
                int a, b;
                cin >> a >> b;
                // 这里还是要检查Find a 和 Find b是不是一样,不然Count会错
                if (Find(a) != Find(b)) {
                    Union(a, b);
                    Count++;
                }
            }
     
            printf("%.2f
    ", Kruskal());
        }
        return 0;
    }


    注意g++交的时候doubl要用f不用lf

  • 相关阅读:
    Docker用途 & 和tomcat的区别
    Ubuntu安装Redis
    Ubuntu查看和设置Root账户
    Oracle常用语句
    Redis知识总结
    Blazor学习笔记01: 使用BootstrapBlazor组件 创建一个具有单表维护功能的表格页面
    NET Core之积沙成塔01: 解决Visual Studio 2019 代码提示为英文
    MySQL系统自带的数据库information schema
    Windows安装mysql方法
    数据库之概念
  • 原文地址:https://www.cnblogs.com/gccbuaa/p/7112117.html
Copyright © 2011-2022 走看看