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,
再从最远的点开始跑, 跑到的最远的的点就是树的直径    

 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;
 8 void read(int &n)
 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];
20 int head[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;
28     edge[num].nxt=head[x];
29     head[x]=num++;
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     }
41     for(int i=head[u];i!=-1;i=edge[i].nxt)
42     {
43         if(edge[i].v!=fa)
44             dfs(edge[i].v,u,now+edge[i].w);
45     }
46 }
47 int  main()
48 {
49      read(n);read(m);
50      memset(head,-1,sizeof(head));
51      for(int i=1;i<=m;i++)
52      {
53          int x,y,z;char c;
54          read(x);read(y);read(z);
55          cin>>c;
56          add_edge(x,y,z);
57          add_edge(y,x,z);
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[500100];
11 int head[MAXN],dp[MAXN];
12 bool vis[MAXN];
13 int n,m,tot,ans;
14 char s[5];
15 
16 inline void init()
17 {
18     memset(vis,false,sizeof(vis));
19     memset(dp,0,sizeof(dp));
20     memset(head,0,sizeof(head));
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];
27     head[u] = tot;
28     e[++tot].to = u;e[tot].w = w;e[tot].nxt = head[v];
29     head[v] = tot;
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);
53             add_edge(x,y,z);
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. 

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。

Output

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

Sample Output

52
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