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. git branch 重命名

    有时候你会有重命名一个git branch的冲动,不要怀疑,这是真的.command bellow will give u a big help,no thanks~ git branch - m o ...

  2. 自定义函数,根据p个数,自适应剧中效果

    //最后投保进程line-height 自适应居中; function self_adaption(){ $('.last_place').each(function(){ var _this = $ ...

  3. UVA 1638 Pole Arrangement

    https://vjudge.net/problem/UVA-1638 题意: n根长度分别为1,2,3,4……n的木棍 将这些木棍竖着排成一列 问从左边看能看到L根,从右边看能看到R根的方案数 将木 ...

  4. 洛谷 P1722 矩阵 II

    题目背景 usqwedf 改编系列题. 题目描述 如果你在百忙之中抽空看题,请自动跳到第六行. 众所周知,在中国古代算筹中,红为正,黑为负…… 给定一个1*(2n)的矩阵(usqwedf:这不是一个2 ...

  5. centos6.8使用脚本一键搭建apache+svn服务

    服务器环境: 脚本如下: #!/bin/bash yum install wget -y mv /etc/yum.repos.d/*.repo /tmp wget -O /etc/yum.repos. ...

  6. u3d局域网游戏网络(c# socket select 模型)

    之前写了一篇. 发完之后第二天实际应用到游戏之后还是发现了一些小毛病. 比如网络模块有重复使用(多对象)的情况.所以将静态类该成了普通类. 比如安卓下会有些异常出现导致游戏逻辑不正常.所以网络相关的函 ...

  7. 【BZOJ1072】【SCOI2007】排列 [状压DP]

    排列 Time Limit: 10 Sec  Memory Limit: 128 MB[Submit][Status][Discuss] Description 给一个数字串s和正整数d, 统计s有多 ...

  8. CRB and Candies(组合数学+求逆元+lcm)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5407 题目: Problem Description CRB has N different cand ...

  9. hadoop2.4.1伪分布式环境搭建

    注意:所有的安装用普通哟用户安装,所以首先使普通用户可以以sudo执行一些命令: 0.虚拟机中前期的网络配置参考: http://www.cnblogs.com/qlqwjy/p/7783253.ht ...

  10. Vue组件-使用插槽分发内容

    在使用组件时,我们常常要像这样组合它们: <app> <app-header></app-header> <app-footer></app-fo ...