hdu 5441 Travel 离线带权并查集

Travel

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5441

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

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

HINT

 

题意

给你一个图，无向边。

有5000个询问

每次询问，问你有多少对城市之间的最长度不超过D

题解：

离线处理，离线之后用并查集维护就好了

按照边权排序，每次就把小于d还没有添加过的边连进去就好了

每次都由并查集维护 ，带权并查集

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <map>
#include <queue>
#include <iomanip>
#include <string>
#include <ctime>
#include <list>
#include <bitset>
typedef unsigned char byte;
#define pb push_back
#define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0)
#define local freopen("in.txt","r",stdin)
#define pi acos(-1)

using namespace std;
const int maxn = 2e4 + 500;
int set[maxn] , n , m , q , ans[maxn] , number[maxn] , cot  = 0;

struct Edge
{
    int u , v , w;
    Edge(int u ,int v ,int w) : u(u) , v(v) , w(w) {}    
    friend bool operator < (const Edge & x,const Edge & y)
    {
        return x.w < y.w;
    }
};

struct Query
{
    int val , idx;
    friend bool operator < (const Query & x,const Query & y)
    {
        return x.val < y.val;
    }    
};

vector<Edge>e;
Query qq[6000];


int find_set(int u)
{
    return set[u] ^ u ? set[u] = find_set(set[u]) : u;
}

int union_set(int u,int v)
{
    int p1 = find_set(u);
    int p2 = find_set(v);
    if(p1 != p2)
    {
        int x = number[p1];
        int y = number[p2];
        cot -= x*(x-1);
        cot -= y*(y-1);
        number[p2] += number[p1];
        cot += (number[p2])*(number[p2]-1);
        number[p1] = 0;
        set[p1] = p2;
    }
}







void initiation()
{
    scanf("%d%d%d",&n,&m,&q);
    for(int i = 1 ; i <= n ; ++ i)
    {
        set[i] = i;
        number[i] = 1;
    }
    for(int i = 1 ; i <= m ; ++ i)
    {
        int u , v , w;
        scanf("%d%d%d",&u,&v,&w);
        e.push_back(Edge(u,v,w));
    }
    sort(e.begin(),e.end());
    for(int i = 1 ; i <= q ; ++ i)
    {
        int d;
        scanf("%d",&d);    
        qq[i].val = d , qq[i].idx = i;
    }
    sort( qq + 1 , qq + 1 + q);
}


void solve()
{
    int ptr = 0;
    for(int i = 1 ; i <= q; ++ i)
    {
        while(ptr < m && e[ptr].w <= qq[i].val)
        {
            union_set(e[ptr].u,e[ptr].v);
            ptr++;
        }
        ans[qq[i].idx] = cot;
    }
    for(int i = 1 ; i <= q; ++ i) printf("%d\n",ans[i]);
}


void clear_all()
{
    e.clear();
    cot = 0;
}


int main(int argc,char *argv[])
{
    int Case;
    scanf("%d",&Case);
    while(Case--)
    {
        initiation();
        solve();
        clear_all();
    }
    return 0;
}