BZOJ 3363 POJ 1985 Cow Marathon 树的直径

题目大意：给出一棵树。求两点间的最长距离。

思路：裸地树的直径。两次BFS，第一次随便找一个点宽搜。然后用上次宽搜时最远的点在宽搜。得到的最长距离就是树的直径。

CODE：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 80010
using namespace std;

int points,edges;
int head[MAX],total;
int next[MAX],aim[MAX],length[MAX];

int f[MAX];

char s[10];

inline void Add(int x,int y,int len);
inline void BFS(int start);

int main()
{
	cin >> points >> edges;
	for(int x,y,len,i = 1;i <= edges; ++i) {
		scanf("%d%d%d%s",&x,&y,&len,s);
		Add(x,y,len),Add(y,x,len);
	}
	BFS(1);
	int _max = 0;
	for(int i = 1;i <= points; ++i)
		if(f[i] > f[_max])
			_max = i;
	BFS(_max);
	int ans = 0;
	for(int i = 1;i <= points; ++i)
		if(f[i] > ans)
			ans = f[i];
	cout << ans << endl;
	return 0;
}

inline void Add(int x,int y,int len)
{
	next[++total] = head[x];
	aim[total] = y;
	length[total] = len;
	head[x] = total;
}

inline void BFS(int start)
{
	static queue<int> q;
	while(!q.empty())	q.pop();
	memset(f,0x3f,sizeof(f));
	f[0] = f[start] = 0;
	q.push(start);
	while(!q.empty()) {
		int x = q.front(); q.pop();
		for(int i = head[x];i;i = next[i])
			if(f[aim[i]] > f[x] + length[i]) {
				f[aim[i]] = f[x] + length[i];
				q.push(aim[i]);
			}
	}
}