zoukankan      html  css  js  c++  java
  • poj 1985 Cow Marathon 树的直径

    题目链接:http://poj.org/problem?id=1985

    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. 

    题意描述:题意很简单,就是求解两个农场之间最远的距离是多少。

    算法分析:树的直径。详细分析:hdu4607

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<cstdlib>
     5 #include<cmath>
     6 #include<algorithm>
     7 #include<queue>
     8 #define inf 0x7fffffff
     9 using namespace std;
    10 const int maxn=40000+10;
    11 const int M = 100000+10;
    12 
    13 int n,m;
    14 struct node
    15 {
    16     int v,w;
    17     int next;
    18 }edge[M*3];
    19 int head[maxn],edgenum;
    20 
    21 void add(int u,int v,int w)
    22 {
    23     edge[edgenum].v=v ;edge[edgenum].w=w ;
    24     edge[edgenum].next=head[u] ;head[u]=edgenum++ ;
    25 
    26     edge[edgenum].v=u ;edge[edgenum].w=w ;
    27     edge[edgenum].next=head[v] ;head[v]=edgenum++ ;
    28 }
    29 
    30 int d[maxn],vis[maxn];
    31 int bfs(int from)
    32 {
    33     memset(d,-1,sizeof(d));
    34     memset(vis,0,sizeof(vis));
    35     queue<int> Q;
    36     Q.push(from);
    37     vis[from]=1 ;d[from]=0 ;
    38     int maxlen=-1,k=0;
    39     while (!Q.empty())
    40     {
    41         int u=Q.front() ;Q.pop() ;
    42         for (int i=head[u] ;i!=-1 ;i=edge[i].next)
    43         {
    44             int v=edge[i].v;
    45             if (!vis[v])
    46             {
    47                 vis[v]=1;
    48                 d[v]=d[u]+edge[i].w;
    49                 Q.push(v);
    50                 if (d[v]>maxlen)
    51                 {
    52                     maxlen=d[v];
    53                     k=v;
    54                 }
    55             }
    56         }
    57     }
    58     return k;
    59 }
    60 
    61 int main()
    62 {
    63     while (scanf("%d%d",&n,&m)!=EOF)
    64     {
    65         memset(head,-1,sizeof(head));
    66         edgenum=0;
    67         int a,b,c;
    68         char str[3];
    69         for (int i=1 ;i<=m ;i++)
    70         {
    71             scanf("%d%d%d%s",&a,&b,&c,str);
    72             add(a,b,c);
    73         }
    74         int v=bfs(1);
    75         int u=bfs(v);
    76         printf("%d
    ",d[u]);
    77     }
    78     return 0;
    79 }
  • 相关阅读:
    C++STL中的unique函数解析
    STL中erase()的用法
    刷题技巧——简易哈希表的实现
    经典面试题目——找到第n个丑数(参考《剑指offer(第二版)》面试题49)
    C++中sort函数小结
    谈谈交叉验证法(个人小结)
    数字序列中某一位数字(《剑指offer》面试题44)
    求1~n整数中1出现的次数(《剑指offer》面试题43)
    2018年美团春招(第二批)题解
    C/C++中字符串和数字互转小结
  • 原文地址:https://www.cnblogs.com/huangxf/p/4367046.html
Copyright © 2011-2022 走看看