poj1985 Cow Marathon（求树的直径）

最新推荐文章于 2019-07-28 10:55:54 发布

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本文介绍了一个基于图论的算法问题——寻找农场间的最长路径，用于规划牛马拉松路线。通过两次广度优先搜索（BFS），首先找到网络中最远的节点，然后从该节点出发再次进行搜索，以确定最长路径的确切距离。

Link：http://poj.org/problem?id=1985

Cow Marathon

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 4351		Accepted: 2178
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

AC code：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<queue>
#define LL long long
using namespace std;
int n,m;
struct node{
    int from;
    int to;
    int w;
    int next;
}Edge[100011*4];
int head[100011];
int tot;
void init()
{
    memset(head,-1,sizeof(head));
    tot=0;
}
void add(int f,int t,int w)
{
    Edge[tot].from=f;
    Edge[tot].to=t;
    Edge[tot].w=w;
    Edge[tot].next=head[f];
    head[f]=tot++;
}
int dis[100011];
int vis[100011];
LL maxx;
int maxx_num;
void bfs(int st)
{
    queue<int>q;
    memset(vis,0,sizeof(vis));
    memset(dis,0,sizeof(dis));
    int cur,nex;
    q.push(st);
    vis[st]=1;
    dis[st]=0;
    maxx=0;
    maxx_num=st;
    while(!q.empty())
    {
        cur=q.front();
        q.pop();
        for(int i=head[cur];i!=-1;i=Edge[i].next)
        {
            nex=Edge[i].to;
            if(!vis[nex])
            {
                vis[nex]=1;
                dis[nex]=dis[cur]+Edge[i].w;
                if(maxx<dis[nex])
                {
                    maxx=dis[nex];
                    maxx_num=nex;
                }
                q.push(nex);
            }
        }
    }
}
int main()
{
    int u,v,w;
    char s[5];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        while(m--)
        {
            scanf("%d%d%d%s",&u,&v,&w,&s);
            add(u,v,w);
            add(v,u,w);
        }
        bfs(1);
        bfs(maxx_num);
        printf("%lld\n",maxx);
    }
    return 0;
}