hdu 5441 Travel 排序并查集_反正5441-优快云博客

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这篇博客主要介绍了如何解决HDU 5441 Travel问题，重点探讨了在解决过程中排序和并查集的数据结构应用，适合对算法和数据结构感兴趣的读者深入理解。

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Travel

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2149 Accepted Submission(s): 740

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

m bidirectional 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

Sample Output

Source

2015 ACM/ICPC Asia Regional Changchun Online

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
#define maxn 1000007
struct Node{
    int u,v,d;
    bool operator <(const Node & a)const{
        return d < a.d;
    }
};
Node edge[maxn];
Node query[maxn];
int ans[maxn];
int fa[maxn],num[maxn];
int find(int u){
    if(fa[u] == u) return u;
    return fa[u]=find(fa[u]);
}

int main(){
    int t,n,m,q;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&n,&m,&q);
        for(int i = 0;i < m; i++)
            scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].d);
        for(int i = 0;i < q; i++){
            scanf("%d",&query[i].d);
            query[i].u = i;
        }
        sort(edge,edge+m);
        sort(query,query+q);
        for(int i = 1;i <= n; i++)
            fa[i] = i, num[i] = 1;
        int f1,f2,cnt=0,res=0;
        for(int i = 0;i < q; i++){
            while(cnt < m && edge[cnt].d <= query[i].d){
                f1 = find(edge[cnt].u);
                f2 = find(edge[cnt].v);
                if(f1 != f2){
                    res += 2*num[f1]*num[f2];
                    fa[f1] = f2;
                    num[f2] += num[f1];
                }
                cnt++;
            }
            ans[query[i].u] = res;
        }
        for(int i = 0;i < q; i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}