POJ 1985 Cow Marathon （求树的直径）

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. 

 

第一行输入n,m表示n个节点，m种关系，然后m行输入三个数a，b，c，表示a，b节点间的距离是c，求节点间最远的距离。S，N，W不影响结果，只在开始时输入就行。

#include<cstdio>
#include<queue>
#include<string.h>
#define M 100000
using namespace std;
int m,ans,flag[M],sum[M],n,a,b,c,i,head[M],num,node;
struct stu
{
    int from,to,val,next;
}st[M];
void init()
{
    num=0;
    memset(head,-1,sizeof(head));
}
void add_edge(int u,int v,int w)
{
    st[num].from=u;
    st[num].to=v;
    st[num].val=w;
    st[num].next=head[u];
    head[u]=num++;
}
void bfs(int fir)
{    
    ans=0;
    int u;
    memset(sum,0,sizeof(sum));
    memset(flag,0,sizeof(flag));
    queue<int>que;
    que.push(fir);
    flag[fir]=1;
    while(!que.empty())
    {
    
        u=que.front();
        que.pop();
        for(i = head[u] ; i != -1 ; i=st[i].next)
        {
            if(!flag[st[i].to] && sum[st[i].to] < sum[u]+st[i].val)
            {
                sum[st[i].to]=sum[u]+st[i].val;
                if(ans < sum[st[i].to])
                {
                    ans=sum[st[i].to];
                    node=st[i].to;
                }
                flag[st[i].to]=1;
                que.push(st[i].to);
            }
        }
    }
}
int main()
{
    char d;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        
        init();
        for(i = 0 ; i < m ; i++)
        {
            scanf("%d %d %d %c",&a,&b,&c,&d);
            add_edge(a,b,c);
            add_edge(b,a,c);
        }
        bfs(1);
        bfs(node);
        printf("%d
",ans);
    }
 }