HDU 5441 Travel （离线处理+并查集）_jack likes to travel around the world, but he does-优快云博客

Travel

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2914 Accepted Submission(s): 998

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

2
6
12

题意：给你一个n个点m条边的无向图，问有多少个点对（a，b）满足从点a到点b的路长不超过d（不超过的意思是每一条路都不超过，但是总的路长可能超过）

思路：将要查询的值排序保存起来，然后按顺序查找，利用并查集，若某条边的两个点祖先不同，则增加的点对数就是两个点集合数目的乘积的两倍。

#include<cstdio>
#include<cstring>
#include<vector>
#include<iostream>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const int maxn = 1e5 + 10;
int pre[maxn], num[maxn];
ll ans[maxn];

struct Edge
{
    int from, to, value, color;
    Edge(int from = 0, int to = 0, int value = 0) : from(from), to(to), value(value){ color = 0; }
    void read()
    {
        scanf("%d%d%d",&from,&to,&value);
    }
    bool operator < (const Edge& b) const
    {
        return value < b.value;
    }
}edge[maxn];

struct Out
{
    int x, id;
    bool operator < (const Out& b) const
    {
        return x < b.x;
    }
}out[maxn];

int find(int x)
{
    return pre[x] == x ? x : pre[x] = find(pre[x]);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--) {
        int n, m, q;
        scanf("%d%d%d",&n,&m,&q);
        for (int i = 0; i < m; i++) edge[i].read();
        for (int i = 0; i < q; i++) {
            scanf("%d",&out[i].x);
            out[i].id = i;
        }
        sort(edge, edge + m);
        sort(out, out + q);
        for (int i = 1; i <= n; i++) pre[i] = i, num[i] = 1;
        int j = 0;
        ll cnt = 0;
        for (int i = 0; i < q; i++) {
            while (j < m && edge[j].value <= out[i].x) {
                int x = edge[j].from;
                int y = edge[j].to;
                int fx = find(x), fy = find(y);
                if (fx != fy) {
                    cnt += (ll)num[fx] * num[fy] * 2;
                    pre[fy] = fx;
                    num[fx] += num[fy];
                }

                j++;
            }
            ans[out[i].id] = cnt;
        }
        for (int i = 0; i < q; i++)
            printf("%d\n",ans[i]);
    }
}