names total mathjax cit not com scrip std php

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

Recommend hujie | We have carefully selected several similar problems for you: 6361 6360 6359 6358 6357 題意:t組數據，每次n個點,m條邊，q次詢問,每次給出一個值，求用到所有邊權不大於這個值的邊的情況下，能夠互相到達的點對的個數 思路:離線並查集，按權值排序即可

#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:摸魚怪的博客分享，歡迎感謝各路大牛的指點~

HDU-5441 Travel 離線-並查集