HDU---5441-Travel（并查集）（2015 Changchun）

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 aren cities
and mbidirectional
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 integerT,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 ncities
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

#include <iostream>
#include <cstring>
#include <vector>
#include <queue>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
#define cin1(x) scanf("%d",&x)
#define cin2(x,y) scanf("%d%d",&x,&y)
#define cin3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define ccout(x) printf("%d\n",x)
int n,m,q,T,sum_val;
int set[20005],sum_set[20005];
struct Road
{
int a,b,val;
Road(int x,int y,int z):a(x),b(y),val(z) {}
friend bool operator<(Road x,Road y)
{
return x.val<y.val;
}
};
struct Question
{
int num,poi;
friend bool operator<(Question x, Question y)
{
return x.poi<y.poi;
}
} que[6005];
int ans[6005];
int set_find(int x)
{
if(set[x]==-1)return x;
return set[x]=set_find(set[x]);
}
void set_friend(int x,int y)
{
x=set_find(x);
y=set_find(y);
if(x!=y)
{
set[x]=y;
sum_val-=sum_set[x]*(sum_set[x]-1)+sum_set[y]*(sum_set[y]-1);
sum_set[y]+=sum_set[x];
sum_set[x]=0;
sum_val+=sum_set[y]*(sum_set[y]-1);
}
}
vector<Road> all_road;
void mem_val()
{
all_road.clear();
for(int i=0; i<=n; ++i)
{
set[i]=-1;
sum_set[i]=1;
}
memset(ans,0,sizeof(ans));
}
int main()
{
cin1(T);
int i,j,k,l,x,y,z;
while(T--)
{
cin3(n,m,q);
mem_val();
for(i=0; i<m; ++i)
{
cin3(x,y,z);
all_road.push_back(Road(x,y,z));
}
for(i=0; i<q; ++i)
{
cin1(x);
que[i].poi=x;
que[i].num=i;
}
sort(all_road.begin(),all_road.end());
sort(que,que+q);

int nowi=0;
sum_val=0;
for(i=0; i<q; ++i)
{
while(nowi<m && all_road[nowi].val<=que[i].poi)
{
set_friend(all_road[nowi].a,all_road[nowi].b);
nowi++;
}
ans[ que[i].num ]=sum_val;
}
for(i=0; i<q; ++i)
{
ccout(ans[i]);
}
}
return 0;
}