Cow Marathon POJ - 1985(树的直径模板题)

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本文介绍了解决 CowMarathon 问题的方法，该问题是寻找地图上两个最远农场之间的路径。通过两次深度优先搜索（DFS），算法找到两对最远农场间的距离，即为马拉松的最长路线。

Cow Marathon POJ - 1985

After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has committed to create a bovine marathon for his cows to run. The marathon route will include a pair of farms and a path comprised of a sequence of roads between them. Since FJ wants the cows to get as much exercise as possible he wants to find the two farms on his map that are the farthest apart from each other (distance being measured in terms of total length of road on the path between the two farms). Help him determine the distances between this farthest pair of farms.

Input
* Lines 1…..: Same input format as “Navigation Nightmare”.
Output
* Line 1: An integer giving the distance between the farthest pair of farms.
Sample Input

Sample Output

Hint
The longest marathon runs from farm 2 via roads 4, 1, 6 and 3 to farm 5 and is of length 20+3+13+9+7=52.

关于树的直径的讲解和证明个人感觉讲的挺明白

code：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N = 1e5+10;
int tot,n,m,nxt[N],dis[N],to[N],head[N],val[N],v,maxd,po;
char c;
void addedge(int u,int v,int w){
    to[++tot] = v;
    val[tot] = w;
    nxt[tot] = head[u];
    head[u] = tot;
}
void dfs(int u,int fa){
    for(int i = head[u]; i; i = nxt[i]){
        v = to[i];
        if(v != fa){
            dis[v] = dis[u] + val[i];
            if(dis[v] > maxd){
                maxd = dis[v];
                po = v;
            }
            dfs(v,u);
        }
    }
}
int main(){
    int x,y,z;
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= m; i++){
        scanf("%d%d%d %c",&x,&y,&z,&c);
        addedge(x,y,z);
        addedge(y,x,z);
    }
    dis[1] = 0;
    dfs(1,0);
    dis[po] = 0;
    dfs(po,0);
    printf("%d",maxd);
    return 0;
}