POJ 1985 Cow Marathon（牛的锻炼，树的直径）

最新推荐文章于 2021-07-28 21:34:28 发布

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本文介绍了一种算法，用于解决寻找地图上两个最远农场间路径的问题。通过构建边和节点，利用广度优先搜索(BFS)两次遍历来找出树的最大直径。

Cow Marathon

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 5000		Accepted: 2429
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

Sample Output

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

题意：

和 POJ 2631（点击查看） 差不多。这题目的大意就是，多让牛锻炼，保护好身体，不生病。

也就是给一些边，问哪两个节点的距离最大。

思路：
树的最大直径，相当于模板题，做了2631，这个就挺简单的了！

代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int MYDD=110300;

int edgenum;//边的数目
int head[MYDD];//节点头指针
void init_edge() {
	edgenum=0;
	memset(head,-1,sizeof(head));
}

struct EDGE { //边的信息
	int u,v,w,next;
} edge[MYDD*2];
void addedge(int u,int v,int w) {//增加边信息
	EDGE Edge= {u,v,w,head[u]};//可以学习这种方法赋值
	edge[edgenum]=Edge;
	head[u]=edgenum++;
}

bool vis[MYDD];//记录节点的访问状态
int dis[MYDD];//记录节点所在直径的最大值
int Tnode,ans;
void BFS(int sx) {
	queue<int> Q;
	memset(vis,false,sizeof(vis));
	memset(dis,0,sizeof(dis));
	vis[sx]=true;
	Q.push(sx);
	Tnode=sx;
	ans=0;
	while(!Q.empty()) {
		int first=Q.front();
		Q.pop();
		for(int j=head[first]; j!=-1; j=edge[j].next) {
			int v=edge[j].v;
			if(!vis[v]) {
				if(dis[v]<edge[j].w+dis[first]) {
					dis[v]=edge[j].w+dis[first];
				}
				vis[v]=true;
				Q.push(v);
				if(ans<dis[v]) {
					ans=dis[v];
					Tnode=v;
				}
			}
		}
	}
}

int main() {
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF) {
		init_edge();
		while(m--) {
			int u,v,w;
			char dd[4];
			scanf("%d%d%d%s",&u,&v,&w,dd);
			addedge(u,v,w);
			addedge(v,u,w);//再写成 u,v,w 试试 
		}
		BFS(1);
		BFS(Tnode);
		printf("%d\n",ans);
	}
}

后：

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