poj:1985:Cow Marathon（求树的直径）

Cow Marathon

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 5496		Accepted: 2685
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. 

 

分析

求树的直径，可以两遍bfs，求树的直径。

我做的是树形dp。

随便找一个节点把无根树变为有根树，考虑随便找一个根，得到一棵有根树。那么每条路径都有一个根。

然后计算其他子节点作根的最长路径。dp[u]表示以u为根的子树中离根的最远距离。则dp[u]=max(dp[v])+1;以u为根的最长路径即为所有u的孩子中，最大的dp值+次大的dp值+1。

code

 #include<cstdio>
 #include<algorithm>
 #include<cstring>

 using namespace std;

 const int MAXN = ;
 struct Edge{
     int to,nxt,w;
 }e[];
 int head[MAXN],dp[MAXN];
 bool vis[MAXN];
 int n,m,tot,ans;
 char s[];

 inline void init()
 {
     memset(vis,false,sizeof(vis));
     memset(dp,,sizeof(dp));
     memset(head,,sizeof(head));
     tot = ;
     ans = ;
 }
 inline void add_edge(int u,int v,int w)
 {
     e[++tot].to = v;e[tot].w = w;e[tot].nxt = head[u];
     head[u] = tot;
     e[++tot].to = u;e[tot].w = w;e[tot].nxt = head[v];
     head[v] = tot;
 }
 void dfs(int u)
 {
     vis[u] = true;
     for (int i=head[u]; i; i=e[i].nxt)
     {
         int v = e[i].to,w = e[i].w;
         if (!vis[v])
         {
             dfs(v);
             ans = max(ans,dp[u]+w+dp[v]);
             dp[u] = max(dp[u],dp[v]+w);
         }
     }
 }
 int main()
 {
     while (~scanf("%d%d",&n,&m))
     {
         init();
         for (int x,y,z,i=; i<=m; ++i)
         {
             scanf("%d%d%d%s",&x,&y,&z,s);
             add_edge(x,y,z);
         }
         dfs();
         printf("%d\n",ans);
     }
     return ;
 }