Roads in the North POJ - 2631
Roads in the North POJ - 2631
Given is an area in the far North comprising a number of villages and roads among them such that any village can be reached by road from any other village. Your job is to find the road distance between the two most remote villages in the area.
The area has up to 10,000 villages connected by road segments. The villages are numbered from 1.
Input
Output
Sample Input
5 1 6
1 4 5
6 3 9
2 6 8
6 1 7
Sample Output
22 题意:找出最长路
题解:树的直径板子
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define ll long long
const int maxn=1e5+;
const int INF=0x3f3f3f3f; struct Edge{
int v,l,next;
}edge[maxn<<];
int vis[maxn],d[maxn],head[maxn],k,node,ans; void init(){
k = ;
memset(head,-,sizeof head);
} void addedge(int u,int v,int l){
edge[k].v = v;
edge[k].l = l;
edge[k].next = head[u];
head[u] = k++; edge[k].v = u;
edge[k].l = l;
edge[k].next = head[v];
head[v] = k++;
}
void dfs(int u,int t){
for(int i=head[u];i!=-;i=edge[i].next)
{
int v = edge[i].v;
if(vis[v] == )
{
vis[v] = ;
d[v] = t + edge[i].l;
if(d[v] > ans)
{
ans = d[v];
node = v;
}
dfs(v,d[v]);
}
}
}
int main()
{
init();
int l,r,len;
while(scanf("%d%d%d",&l,&r,&len) == )
addedge(l,r,len);
memset(vis,,sizeof vis);
vis[] = ;
ans = ;
dfs(,); memset(vis,,sizeof vis);
vis[node] = ;
ans = ;
dfs(node,); printf("%d\n",ans);
};
Roads in the North POJ - 2631的更多相关文章
- poj 2631 Roads in the North
题目连接 http://poj.org/problem?id=2631 Roads in the North Description Building and maintaining roads am ...
- POJ 2631 Roads in the North(树的直径)
POJ 2631 Roads in the North(树的直径) http://poj.org/problem? id=2631 题意: 有一个树结构, 给你树的全部边(u,v,cost), 表示u ...
- poj 2631 Roads in the North【树的直径裸题】
Roads in the North Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 2359 Accepted: 115 ...
- poj 2631 Roads in the North (自由树的直径)
Roads in the North Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4513 Accepted: 215 ...
- POJ 2631 Roads in the North(求树的直径,两次遍历 or 树DP)
题目链接:http://poj.org/problem?id=2631 Description Building and maintaining roads among communities in ...
- Roads in the North(POJ 2631 DFS)
Description Building and maintaining roads among communities in the far North is an expensive busine ...
- 题解报告:poj 2631 Roads in the North(最长链)
Description Building and maintaining roads among communities in the far North is an expensive busine ...
- POJ 2631 Roads in the North (求树的直径)
Description Building and maintaining roads among communities in the far North is an expensive busine ...
- POJ 2631 Roads in the North (模板题)(树的直径)
<题目链接> 题目大意:求一颗带权树上任意两点的最远路径长度. 解题分析: 裸的树的直径,可由树形DP和DFS.BFS求解,下面介绍的是BFS解法. 在树上跑两遍BFS即可,第一遍BFS以 ...
随机推荐
- Http中常见MIME类型
MIME类型 常见MIME类型: 超文本标记语言文本 .html text/html xml文档 .xml text/xml XHTML文档 .xhtml application/xhtml+xml ...
- 打印流-PrintStream
打印流-PrintStream java.io.PrintStream为其他输出流添加了功能,使其他的流能够更方便的打印各种数据值表现形式 PrintStream特点: 1.只负责数据的输入,不负责数 ...
- Day6 盒模型
Day6 盒模型 1.一.标准盒模型(w3c盒模型) 1)组成部分: content + padding + border + margin 内容 ...
- 【java】使用URL和CookieManager爬取页面的验证码和cookie并保存
使用java的net包和io包下的几个工具爬取页面的验证码图片并保存到本地. 然后可以把获取的cookie保存下来,做进一步处理.比如通过识别验证码,进一步使用验证码和用户名,密码,保存下来的cook ...
- svn项目权限控制
[groups] g_manager = zhangsan g_php = lisi g_test = wangwu [/] @g_manager = rw [project:/] @g_manage ...
- jQuery_2_常规选择器-高级选择器2
属性选择器 <a title="num1">num1</a> <a title="num-ad">num2</a> ...
- c++ STL stack容器成员函数
这是后进先出的栈,成员函数比较简单,因为只能操作栈顶的元素.不提供清除什么的函数. 函数 描述 bool s.empty() 栈是否为空(即size=0).若空,返回true,否则,false. vo ...
- Android(java)学习笔记85:使用SQLite的基本流程
- 实现带查询功能的ComboBox控件
实现效果: 知识运用: ComboBox控件的AutoCompleteMode属性 public AutoCompleteMode AutoCompleteMode{get;set;} //属性值为枚 ...
- 第四章 用javascript和DOM去建立一个图片库
把整个图片库的浏览链接集中安排在你的图片库里,只在用户点击了这个主页里的某个图片链接时才把相应的图片传送给它. 代码如下: <body> <ul> <li> < ...