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
 

 

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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\n",ans[i]);
    }
}
View Code

 

 

 

PS:摸鱼怪的博客分享,欢迎感谢各路大牛的指点~
posted @ 2018-08-07 20:59  MengX  阅读(115)  评论(0编辑  收藏  举报

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