Regional Changchun Online--Travel（最小生成树&& 并查集）

Travel

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1852    Accepted Submission(s): 641

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 aren 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

 

Source

2015 ACM/ICPC Asia Regional Changchun Online

 

Recommend

hujie

通过并查集计算新加入的，每次的计算公式为原来的s+合并后的两个集合数量乘积。。。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string>
#include <queue>
#include <string.h>
#include <map>
#include <set>
#include <vector>
#include <algorithm>
#include <stdlib.h>
using namespace std;
#define eps 1e-8
#define INF 20000000005
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rd3(x,y,z) scanf("%d%d%d",&x,&y,&z);
#define rdLL(x) scanf("%I64d",&x)
#define MAXn 20005
#define MAXe 100005
#define QUEnum  5005
struct Node{
   int sta,end,val;
}node[MAXe];

struct Que{
   int queval , num;
}que[QUEnum];

int sett[MAXn];

int sett_find(int x)
{
	if(sett[x]<0)  return x;
	return sett[x]=sett_find(sett[x]);
}


bool cmp1(Node a,Node b){
   return a.val<b.val;
}

bool cmp2(Que a,Que b){
  return a.queval<b.queval;
}
int main ()
{
	int Case;
	rd(Case);
	while(Case--)
	{
		memset(sett,-1,sizeof(sett));
		int sum[MAXn];
		int res[QUEnum];
		for(int i=0;i<MAXn;i++)	sum[i]=1;
		int n,e,quenum;
		rd3(n,e,quenum);
		
		for(int i=0;i<e;i++)  
            rd3(node[i].sta,node[i].end,node[i].val);

		sort(node,node+e,cmp1);  ///这里的排序不是+n 而是e  找了两个小时
		for(int i=0;i<quenum;i++) {
            rd(que[i].queval);
		    que[i].num=i;
		}

		sort(que,que+quenum,cmp2);
		int j=0,s=0;   ///计数有多少能到达
		for(int i=0 ; i<quenum ;i++)  ///不能在这里添加j<e，因为大于的还需要计算
		{
		    while( j<e && node[j].val <= que[i].queval  ){
                int a=node[j].sta,b=node[j].end;

			    int p = sett_find(a);
			    int q = sett_find(b);
			    if(p!=q){
                    s += (sum[q]*sum[p]); ////((sum[q]*(sum[q]+1))>>1) - ((sum[p]*(sum[p]+1))>>1) +
				    sett[q] = p;
				    sum[p]=sum[q]+sum[p]; ///p作为根节点
			    }
			    j++;
		    }
		    res[ que[i].num ] = (s<<1);
		}
		for(int i=0;i<quenum;i++)
            printf("%d\n",res[i]);
	}
	return 0;
}