hdu 5441 Travel(带权并查集)

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 band
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

solution：
有一个n个点的无向图，给出m条边的边权，给出q次询问，每次给出一个值，求用到所有边权不大于这个值的边的情况下，能够互相到达的点对的个数。刚开始以为是最短路变形，但后来发现可以用带权并查集做，把查询读入后从小到大排序，然后对于每次查询，我用并查集维护边权不大于这个值的每个联通块中节点个数，若可以合并u,v，且num[i]代表以i为根的联通块的节点数，则要加上(num[u] + num[v])*(num[u] + num[v] - 1) - num[u] * (num[u] - 1) - num[v] * (num[v] - 1)，因为在次之前我们已经累加过num[i] * (num[i] - 1)。在每次合并时，我们应该选准压缩路径的方向，我每次都是把节点编号大的插在节点编号小的下面。

详细看代码~
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 20500;
int fa[maxn], num[maxn],ans[maxn];
int n;
struct edge{
int u, v, w;
bool operator <(const edge &x)const
{
return w < x.w;
}
}e[maxn*5];
struct Query
{
int id, x;
bool operator <(const Query &a)const
{
return x < a.x;
}
}query[maxn];
void init()
{
for (int i = 1; i <= n; i++)
{
num[i] = 1;
fa[i] = i;
}
}
int find(int x)
{
if (fa[x] == x)return x;
return fa[x] = find(fa[x]);
}
void _union(int x, int y)
{
x = find(x); y = find(y);
if (x > y)swap(x, y);
fa[y] = x;
num[x] += num[y];
}
int main()
{
int t,m,q;
scanf("%d", &t);
while (t--)
{
scanf("%d%d%d", &n, &m, &q);
init();
for (int i = 0; i <m; i++)
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
sort(e, e + m);
for (int i = 0; i < q; i++)
{
query[i].id = i;
scanf("%d", &query[i].x);
}
sort(query, query + q);
int j = 0, tmp = 0;
for (int i = 0; i < q; i++)
{
while (j < m&&e[j].w <= query[i].x)
{
int u = find(e[j].u), v = find(e[j].v);
j++;
if (u == v)continue;
tmp += (num[u] + num[v])*(num[u] + num[v] - 1) - num[u] * (num[u] - 1) - num[v] * (num[v] - 1);
_union(u, v);
}
ans[query[i].id] = tmp;
}
for (int i = 0; i < q; i++)
printf("%d\n", ans[i]);
}
return 0;
}