Roads in the North POJ - 2631(樹的直徑)

Roads in the North

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2992		Accepted: 1477

Description

Building and maintaining roads among communities in the far North is an expensive business. With this in mind, the roads are build such that there is only one route from a village to a village that does not pass through some other village twice. 
Given is an area in the far North comprising a number of villages and roads among them such that any village can be reached by road from any other village. Your job is to find the road distance between the two most remote villages in the area. 

The area has up to 10,000 villages connected by road segments. The villages are numbered from 1. 

Input

Input to the problem is a sequence of lines, each containing three positive integers: the number of a village, the number of a different village, and the length of the road segment connecting the villages in kilometers. All road segments are two-way.

Output

You are to output a single integer: the road distance between the two most remote villages in the area.

Sample Input

5 1 6
1 4 5
6 3 9
2 6 8
6 1 7

Sample Output

22

Source

The UofA Local 1999.10.16

題目大意:給你多個兩點之間的權值，然後求一個最長路

解題思路:求出來樹的直徑就可以了，我一開始想的是權值爲負，求最短路,emmmmm

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
using namespace std;
typedef long long LL;

vector<int> tu[50005],lu[50005];
int check[50005],dis[50005],n,MAXN;
void build(int x,int y,int l)
{
	tu[x].push_back(y);
	lu[x].push_back(l);
}
int dfs(int x)
{
	int tot,k,l,index;
	memset(check,0,sizeof(check));
	memset(dis,0,sizeof(dis));
	tot=0;
	MAXN=0;
	queue<int> qua;
	qua.push(x);
	check[x]=1;
	while(!qua.empty())
	{
		k=qua.front();
		qua.pop();
		for(int i=0;i<tu[k].size();i++)
		{
			index=tu[k][i];
			l=lu[k][i];
			if(check[index])
				continue;
			check[index]=1;
			qua.push(index);
			dis[index]=dis[k]+l;
			if(dis[index]>MAXN)
			{
				MAXN=dis[index];
				tot=index;
			}
		}
	}
	return tot;
}
int main()
{
	int i,x,y,l;
	while(cin>>x>>y>>l)
	{
		build(x,y,l);
		build(y,x,l);
	}
	int t=dfs(1);
	int tot=dfs(t);
	cout<<dis[tot]<<endl;
 return 0;
}