Time Limit: 2000MS   Memory Limit: 30000K
Total Submissions: 7536   Accepted: 3559
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

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. 
 
题意就是求树的直径。水这篇博客为了存板子,自己创的板子,不知道写其他题怎么样,自我感觉很好用,因为自己用着顺手。
 
代码1(BFS版):
 //BFS
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii; const double PI=acos(-1.0);
const double eps=1e-;
const ll mod=1e9+;
const int inf=0x3f3f3f3f;
const int maxn=1e5+;
const int maxm=+;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); int n,m,cnt;//cnt为边数
int dist[maxn],head[maxn];//dist表示最长路,head为存图用的
bool vis[maxn]; struct node{//定义边的结构体
int from,to,val,next;
}edge[maxn<<];//注意是无向图,边数是二倍的 void init()//初始化,不可少
{
cnt=;
memset(head,-,sizeof(head));
} void addedge(int u,int v,int w)
{
edge[cnt].from=u;//起点
edge[cnt].to=v;//终点
edge[cnt].val=w;//权值
edge[cnt].next=head[u];//指向下一条边
head[u]=cnt++;
} int length;//最终的最长路径(树的直径)
int node;//记录端点值 void bfs(int s)
{
queue<int>q;//定义队列
memset(vis,false,sizeof(vis));//初始化,清零
memset(dist,,sizeof(dist));
q.push(s);//入列
vis[s]=true;//记录为遍历过的点
length=;
node=s;
while(!q.empty()){
int u=q.front();
q.pop();
for(int i=head[u];i!=-;i=edge[i].next){//遍历每一条边
int v=edge[i].to;
if(!vis[v]&&dist[v]<dist[u]+edge[i].val){
vis[v]=true;
dist[v]=dist[u]+edge[i].val;//到v的最长路径
if(length<dist[v]){
length=dist[v];//不断更新最长路径
node=v;//更新节点
}
q.push(v);//重新入列,寻找下一个点
}
}
}
} int main()
{
init();
scanf("%d%d",&n,&m);
for(int i=;i<=m;i++){
int u,v,val;char c[];
scanf("%d%d%d%s",&u,&v,&val,c);
addedge(u,v,val);
addedge(v,u,val);
}
bfs();//第一遍找到距离最远的端点
bfs(node);//第二遍找最长距离
printf("%d\n",length);
return ;
}

代码2(DFS vector存图版):

 //DFS-vector存图
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii; const double PI=acos(-1.0);
const double eps=1e-;
const ll mod=1e9+;
const int inf=0x3f3f3f3f;
const int maxn=1e5+;
const int maxm=+;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); struct node{
int to,val;
}; int n,m,length,start;//length为路径长度,start为树的直径开始的节点
vector <node> G[maxn<<];//存图
bool vis[maxn<<];
int dist[maxn<<]; void init()//初始化
{
length=-;
memset(vis,false,sizeof(vis));
for(int i=;i<n;i++)
dist[i]=inf;
} void dfs(int u)
{
vis[u]=true;
for(int i=;i<G[u].size();i++){//对与顶点u相连的点数进行扫描
node v=G[u][i];
if (!vis[v.to]){//如果没有访问过
dist[v.to]=dist[u]+G[u][i].val;//更新节点
if(dist[v.to]>length&&dist[v.to]!=inf){//更新路径长度
length=dist[v.to];
start=v.to;//更新节点
}
dfs(v.to);
}
}
} int main()
{
scanf("%d%d",&n,&m);
for(int i=;i<=m;i++){
int u,v,w;char c[];
scanf("%d%d%d%s",&u,&v,&w,c);
G[u-].push_back({v-,w});
G[v-].push_back({u-,w});
}
init();dist[]=;dfs();
init();dist[start]=;dfs(start);
printf("%d\n",length);
}

代码3(DFS 前向星存图版):

 //DFS-前向星存图
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii; const double PI=acos(-1.0);
const double eps=1e-;
const ll mod=1e9+;
const int inf=0x3f3f3f3f;
const int maxn=1e5+;
const int maxm=+;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); struct node{
int to,nex,val;
}edge[maxn<<]; int n,m,cnt=,length,start;
int dist[maxn<<],head[maxn<<];
bool vis[maxn]; void addedge(int x,int y,int w)
{
edge[++cnt].to=y;
edge[cnt].val=w;
edge[cnt].nex=head[x];
head[x]=cnt;
} void init()//初始化
{
length=-;
memset(vis,false,sizeof(vis));
for(int i=;i<n;i++)
dist[i]=inf;
} void dfs(int u)
{
vis[u]=true;
for(int i=head[u];i;i=edge[i].nex){//对与顶点u相连的点数进行扫描
int v=edge[i].to;
if (!vis[v]){//如果没有访问过
dist[v]=dist[u]+edge[i].val;//更新节点
if(dist[v]>length&&dist[v]!=inf){//更新路径长度
length=dist[v];
start=v;//更新节点
}
dfs(v);
}
}
} int main()
{
scanf("%d%d",&n,&m);
for(int i=;i<=m;i++){
int u,v,w;char c[];
scanf("%d%d%d%s",&u,&v,&w,c);
addedge(u-,v-,w);
addedge(v-,u-,w);
}
init();dist[]=;dfs();
init();dist[start]=;dfs(start);
printf("%d\n",length);
}

OK,滚了。。。

 

POJ 1985.Cow Marathon-树的直径-树的直径模板(BFS、DFS(vector存图)、DFS(前向星存图))的更多相关文章

  1. poj 1985 Cow Marathon

    题目连接 http://poj.org/problem?id=1985 Cow Marathon Description After hearing about the epidemic of obe ...

  2. poj 1985 Cow Marathon【树的直径裸题】

    Cow Marathon Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 4185   Accepted: 2118 Case ...

  3. POJ 1985 Cow Marathon && POJ 1849 Two(树的直径)

    树的直径:树上的最长简单路径. 求解的方法是bfs或者dfs.先找任意一点,bfs或者dfs找出离他最远的那个点,那么这个点一定是该树直径的一个端点,记录下该端点,继续bfs或者dfs出来离他最远的一 ...

  4. poj:1985:Cow Marathon(求树的直径)

    Cow Marathon Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 5496   Accepted: 2685 Case ...

  5. POJ 1985 Cow Marathon (模板题)(树的直径)

    <题目链接> 题目大意: 给定一颗树,求出树的直径. 解题分析:树的直径模板题,以下程序分别用树形DP和两次BFS来求解. 树形DP: #include <cstdio> #i ...

  6. POJ 1985 Cow Marathon (树形DP,树的直径)

    题意:给定一棵树,然后让你找出它的直径,也就是两点中的最远距离. 析:很明显这是一个树上DP,应该有三种方式,分别是两次DFS,两次BFS,和一次DFS,我只写了后两种. 代码如下: 两次BFS: # ...

  7. 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 J ...

  8. POJ 1985 Cow Marathon(树的直径模板)

    http://poj.org/problem?id=1985 题意:给出树,求最远距离. 题意: 树的直径. 树的直径是指树的最长简单路. 求法: 两遍BFS :先任选一个起点BFS找到最长路的终点, ...

  9. 题解报告: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 ge ...

  10. BZOJ 3363 POJ 1985 Cow Marathon 树的直径

    题目大意:给出一棵树.求两点间的最长距离. 思路:裸地树的直径.两次BFS,第一次随便找一个点宽搜.然后用上次宽搜时最远的点在宽搜.得到的最长距离就是树的直径. CODE: #include < ...

随机推荐

  1. mysql的数据库 索引

    1.两种主要的引擎:MyISAM和InnoDB 2.如何查看自己的表是什么类型:http://www.cnblogs.com/luosongchao/archive/2013/05/23/309592 ...

  2. UITableView的代理方法

    一.点击某个cell调用 /** * 点击了第几行调用 */ -(void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NS ...

  3. SSL 证书类型说明: DV OV EV

    内容来自: ssl 证书的三种类型: dv (域名型) , ov (企业型) 和 ev (扩展型) OV.DV和EV证书的区别 另外: 浏览器兼容性测试报告 Symantec 证书为什么相比其他证书要 ...

  4. POJ 2533 Longest Ordered Subsequence LIS O(n*log(n))

    题目链接 最长上升子序列O(n*log(n))的做法,只能用于求长度不能求序列. #include <iostream> #include <algorithm> using ...

  5. Reachability from the Capital(Codeforces Round #490 (Div. 3)+tarjan有向图缩点)

    题目链接:http://codeforces.com/contest/999/problem/E 题目: 题意:给你n个城市,m条单向边,问你需要加多少条边才能使得从首都s出发能到达任意一个城市. 思 ...

  6. 2018 Multi-University Training Contest 1-1002 -Balanced Sequence(括号匹配+贪心)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6299 题目: 题意:t组数据,每组数据给你一个n表示给你n个括号串,这n个括号串之间进行组合,求能够匹 ...

  7. Python的异常处理机制 -- (转)

    当你的程序中出现异常情况时就需要异常处理.比如当你打开一个不存在的文件时.当你的程序中有一些无效的语句时,Python会提示你有错误存在. 下面是一个拼写错误的例子,print写成了Print.Pyt ...

  8. Sqlmap使用教程

    sqlmap也是渗透中常用的一个注入工具,其实在注入工具方面,一个sqlmap就足够用了,只要你用的熟,秒杀各种工具,只是一个便捷性问题,sql注入另一方面就是手工党了,这个就另当别论了. 今天把我一 ...

  9. 【自己练习】linux常见命令——(六)

    菜鸟教程命令大全 http://www.runoob.com/linux/linux-command-manual.html 命令大全:      http://man.linuxde.net/ ta ...

  10. python中的binascii模块

    binascii模块拿来干嘛的? 答:进制转换xxoo #!/usr/bin/env python # encoding:utf-8 # by i3ekr import binascii s = &q ...