HDU 5441 Travel（并查集）

最新推荐文章于 2020-08-11 21:38:03 发布

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本文介绍了一种解决旅行者Jack在有限时间内从一个城市到另一个城市的旅行问题的方法。通过将问题转化为图论中的连通性问题，并采用并查集进行高效处理，实现了对不同时间限制下可行路径数量的有效计算。

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思路：显然的mq这样的做法就T成狗...那么可以给边排个序，询问也排个序，然后扫过去，每次两个连通块合并的时候的答案就是(cnt[u]+cnt[v])*(cnt[u]+cnt[v]-1) - (cnt[u]*(cnt[u]-1))-(cnt[v]*(cnt[v]-1))嘛，然后就做完了

#include<bits/stdc++.h>
using namespace std;
const int maxn = 20000+6;
const int maxm = 100000+7;
int fa[maxn],cnt[maxn],ans[maxn];
int Find(int x){return x==fa[x]?x:fa[x]=Find(fa[x]);}
int n,m,qq;
struct Node
{
	int u,v,w;
}e[maxm];
struct node
{
	int w,id;
}q[maxn];
bool cmp1(node a,node b){return a.w<b.w;}
bool cmp(Node a,Node b){return a.w<b.w;}
void merge(int u,int v)
{
	fa[u]=v;
	cnt[v]+=cnt[u];
}
void init()
{
	for(int i = 0;i<=n;i++)
	{
		fa[i]=i;
		cnt[i]=1;
	}
}
int main()
{
    int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d%d",&n,&m,&qq);
		init();
		for(int i= 1;i<=m;i++)
			scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
		sort(e+1,e+1+m,cmp);
		for(int i = 1;i<=qq;i++)
			scanf("%d",&q[i].w),q[i].id=i;
		sort(q+1,q+1+qq,cmp1);
		int j = 1;
		int res = 0;
        for(int i = 1;i<=qq;i++)
		{
            while(j<=m && e[j].w<=q[i].w)
			{
                int u = Find(e[j].u);
				int v = Find(e[j].v);
				if(u!=v)
				{
					int tmp = cnt[u]+cnt[v];
					res += (tmp*(tmp-1))-(cnt[u]*(cnt[u]-1))-(cnt[v]*(cnt[v]-1));
					merge(u,v);
				}
				j++;
			}
			ans[q[i].id]=res;
		}
		for(int i = 1;i<=qq;i++)
			printf("%d\n",ans[i]);
	}
}

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

cities and

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

minutes, how many pairs of city

are there that Jack can travel from city

without going berserk?

Input

The first line contains one integer

, which represents the number of test case.

For each test case, the first line consists of three integers

and

where

. The Undirected Kingdom has

cities and

bidirectional roads, and there are

queries.

Each of the following

lines consists of three integers

and

where

and

. It takes Jack

minutes to travel from city

to city

and vice versa.

Then

lines follow. Each of them is a query consisting of an integer

where

is the time limit before Jack goes berserk.

Output

You should print

lines for each test case. Each of them contains one integer as the number of pair of cities

which Jack may travel from

within the time limit

.

Note that

and

are counted as different pairs and

and

must be different cities.

Sample Input

Sample Output

2
6
12