HDU 5441 Travel （带权并查集 + 离线操作）

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Travel

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

【思路分析】
参照别人的思路写的。。。这类题一定要多加训练。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int maxn = 2e4 + 50;
const int maxm = 1e5 + 50;
const int maxq = 5e3 + 50;
int n,m,q,sum;
int sets[maxn],ans[maxn],number[maxn];

struct Edge
{
    int u,v,w;
    Edge()
    {

    }
    Edge(int u,int v,int w) : u(u),v(v),w(w)
    {

    }
    bool operator < (const Edge &e) const
    {
        return w < e.w;
    }

}edge[maxm];
struct Query
{
    int val,index;
    bool operator < (const Query &q) const
    {
        return val < q.val;
    }
}query[maxq];

void init()
{
    sum = 0;
    scanf("%d %d %d",&n,&m,&q);
    for(int i = 1;i <= n;i++)
    {
        sets[i] = i;
        number[i] = 1;
    }
    for(int i = 1;i <= m;i++)
    {
        scanf("%d %d %d",&edge[i].u,&edge[i].v,&edge[i].w);
    }
    for(int i = 1;i <= q;i++)
    {
        scanf("%d",&query[i].val);
        query[i].index = i;
    }
    sort(edge + 1,edge + m + 1);
    sort(query + 1,query + q + 1);
}
int finding(int x)
{
    return sets[x] = (sets[x] == x ? x : finding(sets[x]));
}
void merging(int a,int b)
{
    int p1 = finding(a);
    int p2 = finding(b);
    if(p1 != p2)
    {
        int x = number[p1];
        int y = number[p2];
        sum -= x * (x - 1);//即x个结点两两之间边的权值满足条件的点对数
        sum -= y * (y - 1);
        sets[p1] = p2;
        number[p2] += number[p1];
        number[p1] = 0;//以上三步将集合a并入集合b
        sum += (number[p2]) * (number[p2] - 1);
    }
}
void solve()
{
    int num = 1;
    for(int i = 1;i <= q;i++)
    {
        while(num <= m && edge[num].w <= query[i].val)//权值不大于q值时进行合并
        {
            merging(edge[num].u,edge[num].v);
            num++;
        }
        ans[query[i].index] = sum;
    }
    for(int i = 1;i <= q;i++)
        printf("%d\n",ans[i]);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        solve();
    }
    return 0;
}