zoukankan      html  css  js  c++  java
  • HDU_5723_最小生成树+任意两点距离的期望

    Abandoned country

    Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 3449    Accepted Submission(s): 846


    Problem DescriptionAn abandoned country has n(n100000)villages which are numbered from 1 to n . Since abandoned for a long time, the roads need to be re-built. There are m(m1000000) roads to be re-built, the length of each road is wi(wi1000000) . Guaranteed that any two w are different. The roads made all the villages connected directly or indirectly before destroyed. Every road will cost the same value of its length to rebuild. The king wants to use the minimum cost to make all the villages connected with each other directly or indirectly. After the roads are re-built, the king asks a men as messenger. The king will select any two different points as starting point or the destination with the same probability. Now the king asks you to tell him the minimum cost and the minimum expectations length the messenger will walk.
     
    InputThe first line contains an integer T(T10) which indicates the number of test cases. 

    For each test case, the first line contains two integers n,indicate the number of villages and the number of roads to be re-built. Next m lines, each line have three number i,j,w , the length of a road connecting the village i and the village j is w .
     
    Output
    output the minimum cost and minimum Expectations with two decimal places. They separated by a space.
     
    Sample Input
    1
    4 6
    1 2 1
    2 3 2
    3 4 3
    4 1 4
    1 3 5
    2 4 6
     

    题意:求最小生成树,再求任意两点距离的期望。

    最小生成树用kruscal。期望一开始不会求,看题解发现使用dfs:分别求每条边的贡献,一条边的贡献等于这条边的|左子树|*|右子树|*value。

    代码中使用pair和vector

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    using namespace std;
    
    struct Path
    {
        int x,y;
        double value;
    } path[1000005];
    
    typedef pair<int ,int> P;
    vector <P> edge[1000005];
    
    bool cmp(Path a,Path b)
    {
        return a.value<b.value;
    }
    int father[100005],n,m;;
    
    int find(int x)
    {
        if(x!=father[x])
            father[x]=find(father[x]);
        return father[x];
    }
    
    void merge(int a,int b)
    {
        int x=find(a);
        int y=find(b);
        if(x!=y)
            father[x]=y;
    }
    
    long long sumv=0,sumd=0;
    void kruscal()
    {
        for(int i=1; i<=n; i++)
            father[i]=i;
        for(int i=0; i<=n; i++)
            edge[i].clear();
        sort(path,path+m,cmp);
        for(int i=0; i<m; i++)
        {
            int x=path[i].x,y=path[i].y,value=path[i].value;
            if(find(path[i].x)!=find(path[i].y))
            {
                sumv+=path[i].value;
                merge(path[i].x,path[i].y);
                edge[x].push_back(P(y,value));
                edge[y].push_back(P(x,value));
            }
        }
    }
    
    int dfs(int x,int last)
    {
        int cnt=0;
        for(int i=0;i<edge[x].size();i++)
        {
            int y=edge[x][i].first,value=edge[x][i].second;
            if(last!=y)
            {
                int now=dfs(y,x);
                cnt+=now;
                sumd+=1.0*now*(n-now)*value;
            }
        }
        return cnt+1;
    }
    int main() { int t; scanf("%d",&t); while(t--) { sumv=0; sumd=0; scanf("%d%d",&n,&m); for(int i=0; i<m; i++) { scanf("%d%d%lf",&path[i].x,&path[i].y,&path[i].value); } kruscal(); dfs(1,-1); printf("%I64d %.2lf ",sumv,sumd*2.0/(1.0*n)/(n-1.0)); } return 0; }
    Sample Output
    6 3.33
     
    Author
    HIT
     
    Source
  • 相关阅读:
    Html5——视频标签使用
    Android的四大组件
    Android 硬编码
    按键事件处理
    android Keycode 完全对照表
    Activity的生命周期
    音乐播放控制
    Android permission 访问权限大全
    制作留言板相关资料
    adb查询log命令
  • 原文地址:https://www.cnblogs.com/jasonlixuetao/p/5702215.html
Copyright © 2011-2022 走看看