hdu 3938 Portal

 http://acm.hdu.edu.cn/showproblem.php?pid=3938

                            Portal

                                Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                Total Submission(s): 113    Accepted Submission(s): 56

 

Problem Description

ZLGG found a magic theory that the bigger banana the bigger banana peel .This important theory can help him make a portal in our universal. Unfortunately, making a pair of portals will cost min{T} energies. T in a path between point V and point U is the length of the longest edge in the path. There may be lots of paths between two points. Now ZLGG owned L energies and he want to know how many kind of path he could make.

 

Input

There are multiple test cases. The first line of input contains three integer N, M and Q (1 < N ≤ 10,000, 0 < M ≤ 50,000, 0 < Q ≤ 10,000). N is the number of points, M is the number of edges and Q is the number of queries. Each of the next M lines contains three integers a, b, and c (1 ≤ a, b ≤ N, 0 ≤ c ≤ 10^8) describing an edge connecting the point a and b with cost c. Each of the following Q lines contain a single integer L (0 ≤ L ≤ 10^8).

 

Output

Output the answer to each query on a separate line.

 

Sample Input

10 10 10

7 2 1

6 8 3

4 5 8

5 8 2

2 8 9
6 4 5
2 1 5
8 10 5
7 3 7
7 8 8

10

6

1

5

9

1

8

2

7

6

 

Sample Outpu

36

13

1

13

36

1

36

2

16

13

 

半天沒看懂題目，水題一個。

問你有多少對(u,v)使得u，v之間的所有路徑的最大邊權的最小值小於給定的查詢值L。這種題一般都是把查詢和邊統一排序，然後並查集處理。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define M 70050
#define N 10050
struct node
{
	int u,v,c;
}a[M];
int father[N],rank[N];
__int64 V[N],ans[N],coun;
int find(int t)
{
	if(father[t]!=t)
		father[t]=find(father[t]);
	return father[t];
}
void Union(int x,int y)
{
	int a=find(x),b=find(y);
	if(a==b)
		return ;
	coun+=V[a]*V[b];
	if(rank[a]<rank[b])
	{
		father[a]=b;
		V[b]+=V[a];
	}
	else
	{
		father[b]=a;
		V[a]+=V[b];
		if(rank[a]==rank[b])
			rank[a]++;
	}
}
bool cmp(node a,node b)
{
	if(a.c==b.c)
		return a.u>b.u;
	return  a.c<b.c;
}
int main()
{
	int n,m,q,i,u,v,c,L;
	while(scanf("%d%d%d",&n,&m,&q)!=EOF)
	{
		for(i=0;i<m;i++)
		{
			scanf("%d%d%d",&u,&v,&c);
			a[i].u=u;a[i].v=v;a[i].c=c;
		}
		for(i=0;i<q;i++)
		{
			scanf("%d",&L);
			a[m+i].u=-(i+1),a[m+i].v=-1;a[m+i].c=L;
		}
		sort(a,a+m+q,cmp);
		coun=0;
		for(i=1;i<=n;i++)
		{
			father[i]=i;
			rank[i]=1;
			V[i]=1;
		}
		for(i=0;i<m+q;i++)
		{
			u=a[i].u;v=a[i].v;
			if(u>0)
				Union(u,v);
			else
				ans[-u]=coun;
		}
		for(i=1;i<=q;i++)
			printf("%I64d\n",ans[i]);
	}
	return 0;
}