2586 How far away ？

最新推荐文章于 2020-09-23 20:48:39 发布

原创最新推荐文章于 2020-09-23 20:48:39 发布 · 854 阅读

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本文介绍了一种解决乡村中询问从一个房屋到另一个房屋的最短距离问题的方法，利用单源最短路径算法来解答乡村居民的问题。

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Problem Description

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.

Input

First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

Sample Input

Sample Output

这里才用的是单源最短路径，注意在不能到达的时候，输出是0！

#include <iostream>
#include <queue>
using namespace std;
const int size=40001;
const int maxs=2000000000;
struct node 
{
	int a;
	int b;
	int dis;
	node *next;
};
unsigned int dis[size];
node *mynode[size]={0};
bool visited[size]={0};
int start,end;
int main()
{
	for (int j=0;j<size;++j)
	{
		mynode[j]=NULL;
		dis[j]=maxs;
		visited[j]=0;
	}

	queue<node*> q;
	int T,n,m,a,b,c;
	cin>>T;
	while (T>0)
	{
		cin>>n>>m;
		while (n-1>0)
		{
			cin>>a>>b>>c;
			node *na=new node;
			node *nb=new node;
			na->a=nb->b=a;
			na->b=nb->a=b;
			na->dis=nb->dis=c;
			if (mynode[a]==NULL)
			{
				mynode[a]=na;
				na->next=NULL;
			}
			else
			{
				na->next=mynode[a];
				mynode[a]=na;
			}
			if (mynode[b]==NULL)
			{
				mynode[b]=nb;
				nb->next=NULL;
			}
			else
			{
				nb->next=mynode[b];
				mynode[b]=nb;
			}		
			--n;
		}
		while (m>0)
		{
			int sum=-1;
			node *temp;
			cin>>a>>b;
			start=a;
			end=b;
			dis[start]=0;
			temp=mynode[a];
			while (temp!=NULL)
			{
				dis[temp->b]=temp->dis;
				q.push(temp);
				visited[temp->a]=visited[temp->b]=1;
				temp=temp->next;
			}
			while (!q.empty())
			{
				temp=q.front();
				q.pop();
				
				temp=mynode[temp->b];
				while (temp!=NULL)
				{
					if (!visited[temp->b])
					{
						q.push(temp);
						visited[temp->b]=1;
					}
					if (dis[temp->b]>dis[temp->a]+temp->dis)
						dis[temp->b]=dis[temp->a]+temp->dis;
					temp=temp->next;
				}
			}
			if (dis[end]==maxs)
				cout<<0<<endl;
			else cout<<dis[end]<<endl;
			--m;
			for (int j=0;j<size;++j)
			{
				dis[j]=maxs;
				visited[j]=0;
			}
		}
		for (int j=0;j<size;++j)
		{
			node* temp;
			while (mynode[j]!=NULL)
			{
				temp=mynode[j];
				mynode[j]=mynode[j]->next;
				delete temp;
			}
		}
		--T;
	}
	return 0;
}