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   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 
#include 
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 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<next;
				delete temp;
			}
		}
		--T;
	}
	return 0;
}