poj1985 Cow Marathon（樹的直徑）

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Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 9110 Accepted: 4171 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

思路

先搞清樣例輸入：
from,to,val,direction(from到to有一條長val的邊，to在from的direction方向)
這里不用考慮方向，直接建雙向邊。
題意：求解兩個牧場之間最長的距離。
即求樹的直徑即可。
樹的直徑求解步驟：

任意找一點作為起始點，找到離他最遠的一點pos1，根據樹直徑的性質，此時pos1一定是樹直徑的一個端點。

把pos1當做起點，找離pos1最遠的點pos2;
pos1------pos2即為樹的直徑。

方法：

兩遍dfs或者兩遍bfs。

樹形dp

AC代碼：

• 法1：兩遍dfs（bfs略）：
• 時間復雜度O(n)
• 優點：好記錄路徑
• 缺點：寫起來相對復雜一點點

#include <iostream> #include <cstring> #include <string> using namespace std; const int N = 1e5+5; struct Edge {int from;int to;int val;int nxt; }edge[N]; int head[N<<1],idx; void init() {memset(head,-1,sizeof(head));idx = 0; } inline void add_edge(int from,int to,int val) {edge[idx].from = from;edge[idx].to = to;edge[idx].val = val;edge[idx].nxt = head[from];head[from] = idx++; } bool vis[N]; int pos1,pos2,maxx; void dfs(int s,int dis) {if(dis > maxx){pos1 = s;maxx = dis;//cout<<pos1<<endl;}for(int i = head[s]; ~i;i = edge[i].nxt){int to = edge[i].to;int val = edge[i].val;if(!vis[to]){vis[to] = true;dfs(to,dis+val);}}return; } int main() {int n,m;cin>>n>>m;int x,y,v;string d;init();while(m--){cin>>x>>y>>v>>d;add_edge(x,y,v);add_edge(y,x,v);}memset(vis,false,sizeof(vis));maxx = 0;dfs(1,0);//cout<<pos1<<endl;pos2 = pos1;maxx = 0;memset(vis,false,sizeof(vis));dfs(pos1,0);//cout<<pos1<<endl;cout<<maxx<<endl;return 0; }

簡化代碼：

#include <iostream> #include <cstring> #include <string> using namespace std; const int N = 1e5+5; struct Edge {int from;int to;int val;int nxt; } edge[N]; int head[N<<1],idx; void init() {memset(head,-1,sizeof(head));idx = 0; } inline void add_edge(int from,int to,int val) {edge[idx].from = from;edge[idx].to = to;edge[idx].val = val;edge[idx].nxt = head[from];head[from] = idx++; } int pos1,pos2,maxx; void dfs(int u,int fa,int dis) {if(dis > maxx){pos1 = u;maxx = dis;}for(int i = head[u]; ~i; i = edge[i].nxt){int to = edge[i].to;int val = edge[i].val;if(to != fa){dfs(to,u,dis+val);}} } int main() {int n,m;cin>>n>>m;int x,y,v;string d;init();while(m--){cin>>x>>y>>v>>d;add_edge(x,y,v);add_edge(y,x,v);}dfs(1,0,0);pos2 = pos1;dfs(pos1,0,0);cout<<maxx<<endl;return 0; }

• 法2：dp
• 時間復雜度O(n)
• 優點：寫起來簡單
• 缺點：不好記錄路徑

#include <iostream> #include <cstring> #include <string> using namespace std; const int N = 1e5+5; struct Edge {int from;int to;int val;int nxt; } edge[N]; int head[N<<1],idx; void init() {memset(head,-1,sizeof(head));idx = 0; } inline void add_edge(int from,int to,int val) {edge[idx].from = from;edge[idx].to = to;edge[idx].val = val;edge[idx].nxt = head[from];head[from] = idx++; } int maxx; int dp[N]; void DP(int u,int fa) {for(int i = head[u]; ~i; i = edge[i].nxt){int to = edge[i].to;int val = edge[i].val;if(to != fa){DP(to,u);maxx = max(maxx,dp[u]+dp[to]+val);dp[u] = max(dp[u],dp[to]+val);}}return; } int main() {int n,m;cin>>n>>m;int x,y,v;string d;init();while(m--){cin>>x>>y>>v>>d;add_edge(x,y,v);add_edge(y,x,v);}DP(1,0);cout<<maxx<<endl;return 0; }

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