POJ 1985 Cow Marathon

    此题求的是树的最长链， 要注意此题不能用dis[maxn][maxn]来记录a, b两点的距离，这样会超空间， 也就是超内存，要用另一种储存方式储存。此题思想很简单， 随便选一个点，搜此点能到的最远距离， 并记录最远距离的点， 然后再从最远距离的点搜这点能到的最远距离， 这个dis就是答案。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#define mem(a) memset(a, 0, sizeof(a))
using namespace std;int m, n, ans, flag, num, head[100005];
bool vis[100005];struct eage
{int to;int next;int eg;
}e[100005];void add(int a, int b, int eg)
{e[num].to = b;e[num].eg = eg;e[num].next = head[a];head[a] = num++;e[num].to = a;e[num].eg = eg;e[num].next = head[b];head[b] = num++;
}void bfs(int a)
{int dis = 0, he, to;queue<int> q;q.push(a);q.push(dis);while(!q.empty()){he = q.front();vis[he] = 1;q.pop();dis = q.front();q.pop();to = head[he];while(vis[e[to].to] == 1&&to != -1){to = e[to].next;}while(to != -1&&vis[e[to].to] == 0){q.push(e[to].to);q.push(dis + e[to].eg);if(ans < (dis + e[to].eg))ans = dis + e[to].eg, flag = e[to].to;to = e[to].next;while(vis[e[to].to] == 1&&to != -1){to = e[to].next;}}}
}int main(int argc, char *argv[])
{int i, j, k, l;ans = 0;num = 0;mem(e);mem(vis);memset(head, -1, sizeof(head));scanf("%d%d",&m, &n);for(i = 0;i < n;i++){scanf("%d%d%d%*c%*c%*c", &j, &k, &l);add(j, k, l);}bfs(1);mem(vis);bfs(flag);printf("%d\n", ans);return 0;
}

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