2586 How far away ？

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

2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1

 

Sample Output

10
25
100
100

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;
}