## Input

* Lines 1…..: Same input format as “Navigation Nightmare”.

## POJ 1985 Cow Marathon(树的直径)，pojmarathon

Cow Marathon

Time Limit: 2000MS

Memory Limit: 30000K

Total Submissions: 5357

Accepted: 2630

Case 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 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 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.

Source

USACO 2004 February     先假设节点1是根节点， 跑到离他最远的点，

`````` 1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 #define lli long long int
6 using namespace std;
7 const int MAXN=100001;
9 {
10     char c='+';int x=0;bool flag=0;
11     while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}
12     while(c>='0'&&c<='9')
13     x=(x<<1)+(x<<3)+c-48,c=getchar();
14     flag==1?n=-x:n=x;
15 }
16 struct node
17 {
18     int u,v,w,nxt;
19 }edge[MAXN];
21 int num=1;
22 int n,m;
23 void add_edge(int x,int y,int z)
24 {
25     edge[num].u=x;
26     edge[num].v=y;
27     edge[num].w=z;
30 }
31 int dp[MAXN];
32 int ans=0;
33 int ed=0;
34 void dfs(int u,int fa,int now)
35 {
36     if(now>ans)
37     {
38         ans=now;
39         ed=u;
40     }
42     {
43         if(edge[i].v!=fa)
44             dfs(edge[i].v,u,now+edge[i].w);
45     }
46 }
47 int  main()
48 {
51      for(int i=1;i<=m;i++)
52      {
53          int x,y,z;char c;
55          cin>>c;
58      }
59      dfs(1,0,0);
60      dfs(ed,0,0);
61      printf("%d",ans);
62     return 0;
63 }
``````

http://www.bkjia.com/cjjc/1219082.htmlwww.bkjia.comtruehttp://www.bkjia.com/cjjc/1219082.htmlTechArticlePOJ 1985 Cow Marathon(树的直径)，pojmarathon Cow
Marathon Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 5357
Accepted: 2630 Case Time Limit: 1000MS Description Af…

## 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.

# Cow Marathon

Time Limit: 2000MS

Memory Limit: 30000K

Total Submissions: 5496

Accepted: 2685

Case Time Limit: 1000MS

## code

`````` 1 #include<cstdio>
2 #include<algorithm>
3 #include<cstring>
4
5 using namespace std;
6
7 const int MAXN = 50010;
8 struct Edge{
9     int to,nxt,w;
10 }e;
12 bool vis[MAXN];
13 int n,m,tot,ans;
14 char s;
15
16 inline void init()
17 {
18     memset(vis,false,sizeof(vis));
19     memset(dp,0,sizeof(dp));
21     tot = 0;
22     ans = 0;
23 }
24 inline void add_edge(int u,int v,int w)
25 {
26     e[++tot].to = v;e[tot].w = w;e[tot].nxt = head[u];
28     e[++tot].to = u;e[tot].w = w;e[tot].nxt = head[v];
30 }
31 void dfs(int u)
32 {
33     vis[u] = true;
34     for (int i=head[u]; i; i=e[i].nxt)
35     {
36         int v = e[i].to,w = e[i].w;
37         if (!vis[v])
38         {
39             dfs(v);
40             ans = max(ans,dp[u]+w+dp[v]);
41             dp[u] = max(dp[u],dp[v]+w);
42         }
43     }
44 }
45 int main()
46 {
47     while (~scanf("%d%d",&n,&m))
48     {
49         init();
50         for (int x,y,z,i=1; i<=m; ++i)
51         {
52             scanf("%d%d%d%s",&x,&y,&z,s);
54         }
55         dfs(1);
56         printf("%d\n",ans);
57     }
58     return 0;
59 }
``````

## Sample Input

``````7 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 7 2 S
``````

## 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.

## Output

* Line 1: An integer giving the distance between the farthest pair of
farms.

## Sample Output

``````52
`````` 