Cow Marathon（两次dfs求树的直径）

http://poj.org/problem?id=1985

Cow Marathon

Time Limit: 2000MS Memory Limit: 30000KTotal Submissions: 4698 Accepted: 2323Case Time Limit: 1000MS

Description

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

7 61 6 13 E6 3 9 E3 5 7 S4 1 3 N2 4 20 W4 7 2 S

Sample Output

52

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.

题目意思是没有环的，那么就是求树的直径，这题好像是简化了。。。两次dfs。。。随便取一个点（假设是1），dfs找到距离它最远的点u，再从那个最远的点dfs，找到一个新的最远的点v，那么u，v即为树的直径

#include <cstdio>#include <string.h>#include <algorithm>using namespace std;int const MAX = 100000;struct Edge{    int val, to, next;}e[MAX*4];int head[MAX];int n, m, tot;int vis[MAX], ans, f;void addedge(int u, int v, int w){    e[++tot].to = v;    e[tot].val = w;    e[tot].next = head[u];    head[u] = tot;}void dfs(int u,int num){    vis[u]=1;    if(num>ans){        ans=num;        f=u;    }    for(int i=head[u];i!=-1;i=e[i].next)    {        if(!vis[e[i].to])            dfs(e[i].to,num+e[i].val);    }}int main(){    memset(head, -1, sizeof(head));    scanf("%d %d", &n, &m);    tot = 0;    while(m--){        int u, v, w;        char c;        scanf("%d %d %d %c", &u, &v, &w, &c);        addedge(u, v, w);        addedge(v, u, w);    }    ans=0;    memset(vis,0,sizeof vis);    dfs(1,0);    memset(vis,0,sizeof vis);    dfs(f,0);    printf("%d\n",ans);    return 0;}