【POJ1985】Cow Marathon（基础树的直径）

CowMarathon算法解析

最新推荐文章于 2022-07-08 23:01:31 发布

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本文介绍了一种名为CowMarathon的算法问题，旨在寻找农场地图上距离最远的两个农场之间的路径长度。提供了两种实现方案，一种是使用广度优先搜索（BFS），另一种是采用深度优先搜索（DFS）。通过具体示例展示了算法的具体应用。

Cow Marathon

Time Limit:2000MS Memory Limit:30000KB 64bit IO Format:%lld & %llu

Submit Status

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.

第一种：套用的模板

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
int dist[40010];//要足够大，不然就runtime error
bool vis[40010];
int head[40010];
int n,m;
struct Edge
{
	int from,to,val,next;
 } edge[80010];
 int edgenum;
void addedge(int u,int v,int w)
{
edge[edgenum].from=u;
edge[edgenum].to=v;
edge[edgenum].val=w;
edge[edgenum].next=head[u];
head[u]=edgenum++;	
}
void init()
{
	edgenum=0;
	memset(head,-1,sizeof(head));
}
int ans;
int node;
void bfs(int s)
{
	queue<int>q;
	memset(dist,0,sizeof(dist));
	memset(vis,false,sizeof(vis));
	q.push(s);
	vis[s]=true;
	ans=0;
	node=s;
	while(!q.empty())
    {
	int u=q.front();
	q.pop();
	for(int i=head[u];i!=-1;i=edge[i].next)
	{
		int v=edge[i].to;
		if(!vis[v])
		{
		if(dist[v]<dist[u]+edge[i].val)
			{
				dist[v]=dist[u]+edge[i].val;
			}
			vis[v]=true;
			q.push(v);
		}
	  }
   }
   for(int i=1;i<=n;i++)//注意是1到n,不是0到n-1
   {
   	if(ans<dist[i])
   	{
   		ans=dist[i];
   		node=i;
	   }
	   }	
}
int main()
{
	int a,b,c;
	char h[2];
	while(~scanf("%d%d",&n,&m))
	{
		init();
		for(int i=0;i<m;i++)
		{
			scanf("%d%d%d%s",&a,&b,&c,&h);
			addedge(a,b,c);
			addedge(b,a,c);
		}
		bfs(1);
		bfs(node);
		printf("%d\n",ans);
	}
}

第二种：参考的别人代码，看起来好简单，然而本渣渣初学不懂，dfs不太会用(╥╯^╰╥)

链接：http://blog.youkuaiyun.com/sio__five/article/details/19016079

using namespace std;  
struct node  
{  
    int x, w;  
};  
int n, m, mx = 0, k;  
vector<node> v[maxn];  
int vis[maxn] = {0};  
void dfs(int x, int s)  
{  
    if (s > mx) mx = s, k = x;  
    for (int i = 0; i < v[x].size(); i++) if (!vis[v[x][i].x])  
    {  
        vis[v[x][i].x] = 1;  
        dfs(v[x][i].x, s + v[x][i].w);  
        vis[v[x][i].x] = 0;  
    }  
}  
int main ()  
{  
    while(scanf("%d%d", &n, &m) != EOF)  
    {  
        int x, y;  
        node tmp;  
        char ch;  
        for (int i = 0; i <= n; i++) v[i].clear();  
        while(m--)  
        {  
            scanf("%d%d%d %c", &x, &y, &tmp.w, &ch);  
            tmp.x = y, v[x].push_back(tmp);  
            tmp.x = x, v[y].push_back(tmp);  
        }  
        mx = 0;  
        mem(vis, 0), vis[1] = 1;  
        dfs(1, 0);  
        mem(vis, 0), vis[k] = 1;  
        dfs(k, 0);  
        cout<<mx<<endl;  
    }  
    return 0;  
}