HDU5441(2015长春网赛)——Travel(并查集应用,离线处理)

Travel

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3283    Accepted Submission(s): 1117

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

 

题意：简单来说就是给你一张图。每次查询的时候给你一个值，能走的边权值不能超过那个值。求连通点对。

思路：就是根据权值对边排序，并查集判断连通性。算一下就行。但是注意不能在线处理，得先把所有的查询读进去，然后排个序，这样可以保证合并操作不会超过n次。复杂度是nlogn。

#include <cmath>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <ctime>
#include <cstdlib>
#include <iostream>
using namespace std;
#define MAXN 20010
#define MAXM 100010
#define LEN 200010
#define INF 1e9+7
#define MODE 1000000
#define pi acos(-1)
#define g 9.8
typedef long long ll;
struct edge{
    int u,v,w;
};
int cmp(edge &a,edge &b){
    return a.w<b.w;
}
edge e[MAXM];

int par[MAXN];
int r[MAXN];
void init(int n){
    for(int i=1;i<=n;i++){
        par[i]=i;
        r[i]=1;
    }
}

int find(int x){
    if(par[x]==x)
        return x;
    else
        return par[x]=find(par[x]);
}

void unite(int x,int y){
    x=find(x);
    y=find(y);
    if(x==y)
        return;
    par[y]=x;
    r[x]+=r[y];
}
struct o{
    int num,x;
    long long res;
};
o op[5010];
int cmp1(o a,o b){
    return a.x<b.x;
}
int cmp2(o a,o b){
    return a.num<b.num;
}
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++){
            scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
        }
        sort(e,e+m,cmp);
        for(int i=0;i<q;i++){
            scanf("%d",&op[i].x);
            op[i].num=i;
        }
        sort(op,op+q,cmp1);
        init(n);
        int cnt=0;
        for(int i=0;i<q;i++){
            while(e[cnt].w<=op[i].x){
                unite(e[cnt].u, e[cnt].v);
                cnt++;
            }
            long long sum=0;
            for(int i=1;i<=n;i++){
                if(par[i]==i){
                    sum+=(long long)r[i]*((long long)r[i]-1);
                }
            }
            op[i].res=sum;
        }
        sort(op,op+q,cmp2);
        for(int i=0;i<q;i++){
            printf("%lld\n",op[i].res);
        }
    }
}