zoukankan      html  css  js  c++  java
  • HDU-5441 Travel 离线-并查集

    Travel

    Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 4680    Accepted Submission(s): 1532


    Problem Description
    Jack likes to travel around the world, but he doesn’t like to wait. Now, he is traveling in the Undirected Kingdom. There are n cities and m bidirectional roads connecting the cities. Jack hates waiting too long on the bus, but he can rest at every city. Jack can only stand staying on the bus for a limited time and will go berserk after that. Assuming you know the time it takes to go from one city to another and that the time Jack can stand staying on a bus is x minutes, how many pairs of city (a,b) are there that Jack can travel from city a to b without going berserk?
     
    Input
    The first line contains one integer T,T5, which represents the number of test case.

    For each test case, the first line consists of three integers n,m and q where n20000,m100000,q5000. The Undirected Kingdom has n cities and mbidirectional roads, and there are q queries.

    Each of the following m lines consists of three integers a,b and d where a,b{1,...,n} and d100000. It takes Jack d minutes to travel from city a to city b and vice versa.

    Then q lines follow. Each of them is a query consisting of an integer x where x is the time limit before Jack goes berserk.

     
    Output
    You should print q lines for each test case. Each of them contains one integer as the number of pair of cities (a,b) which Jack may travel from a to b within the time limit x.

    Note that (a,b) and (b,a) are counted as different pairs and a and b must be different cities.
     
    Sample Input
    1
    5 5 3
    2 3 6334
    1 5 15724
    3 5 5705
    4 3 12382
    1 3 21726
    6000
    10000
    13000
     
    Sample Output
    2 6 12
     
    Source
     
    Recommend
    hujie   |   We have carefully selected several similar problems for you:  6361 6360 6359 6358 6357 
     
    题意:t组数据,每次n个点,m条边,q次询问,每次给出一个值,求用到所有边权不大于这个值的边的情况下,能够互相到达的点对的个数
    思路:离线并查集,按权值排序即可
    #include<bits/stdc++.h>
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define ll long long
    #define mem(a,b) memset(a,b,sizeof(a))
    using namespace std;
    const int maxn = 20005;
    int pre[maxn],r[maxn];
    ll ans[maxn];
    struct node
    {
        int u,v;
        ll val;
        bool friend operator < (const node xx,const node yy)
        {
            return xx.val<yy.val;
        }
    } e[maxn*10];
    struct pp
    {
        ll dis;
        int id;
        bool friend operator < (const pp xx,const pp yy)
        {
            return xx.dis<yy.dis;
        }
    } Q[5005];
    void init(int n)
    {
        for(int i=1; i<=n; i++)
            {pre[i]=i;r[i]=1;}
    }
    int Find(int x)
    {
        return x==pre[x]?x:pre[x]=Find(pre[x]);
    }
    int main()
    {
        int t,n,m,q;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d %d %d",&n,&m,&q);
            init(n);
            for(int i=1; i<=m; i++)
            {
                scanf("%d %d %lld",&e[i].u,&e[i].v,&e[i].val);
            }
            sort(e+1,e+1+m);
            for(int i=1; i<=q; i++)
            {
                scanf("%lld",&Q[i].dis);
                Q[i].id=i;
            }
            sort(Q+1,Q+1+q);
            int j=1,cnt=0;
            for(int i=1; i<=q; i++)
            {
                while(j<=m&&Q[i].dis>=e[j].val)
                {
    
                    int u=Find(e[j].u);
                    int v=Find(e[j].v);
                    if(u!=v)
                    {
                        cnt+=r[u]*r[v];
                        pre[u]=v;
                        r[v]+=r[u];
                    }
                    j++;
                }
                ans[Q[i].id]=cnt<<1;
            }
            for(int i=1;i<=q;i++)
                printf("%lld
    ",ans[i]);
        }
    }
    View Code
    PS:摸鱼怪的博客分享,欢迎感谢各路大牛的指点~
  • 相关阅读:
    Microsoft Biztalk Server 2000简介
    BizTalk学习笔记系列之二:实例说明如何使用BizTalk
    BizTalk学习笔记系列之三:企业集成应用和BizTalk
    简单状态机Workflow基于Web应用【转】
    C#类、接口、虚方法和抽象方法
    多表查询语句写法、数据库数字如何转化为汉子、Sql语句拼接
    IsPostBack用法
    Net前台页面如何调用后台cs变量
    aspx页面中写if else 语句的方法,
    查询数据库最大的索引、静态类与非静态类的区别、后台操作DIV样式的方法、C#操作TreeView组件中的一些常用方法及具体实现
  • 原文地址:https://www.cnblogs.com/MengX/p/9439425.html
Copyright © 2011-2022 走看看