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,T≤5, which represents the number of test case.

For each test case, the first line consists of three integers n,m and q where n≤20000,m≤100000,q≤5000. 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 d≤100000. 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

2015 ACM/ICPC Asia Regional Changchun Online

 

 

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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]);
    }
}

 

 

 

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