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
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 }