hdu 4123(树形dp+倍增)
Bob’s Race
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3741 Accepted Submission(s): 1206
wants to hold a race to encourage people to do sports. He has got
trouble in choosing the route. There are N houses and N - 1 roads in his
village. Each road connects two houses, and all houses are connected
together. To make the race more interesting, he requires that every
participant must start from a different house and run AS FAR AS POSSIBLE
without passing a road more than once. The distance difference between
the one who runs the longest distance and the one who runs the shortest
distance is called “race difference” by Bob. Bob does not want the “race
difference”to be more than Q. The houses are numbered from 1 to N. Bob
wants that the No. of all starting house must be consecutive. He is now
asking you for help. He wants to know the maximum number of starting
houses he can choose, by other words, the maximum number of people who
can take part in his race.
The first line of each test case contains two integers N and M. N is the number of houses, M is the number of queries.
The
following N-1 lines, each contains three integers, x, y and z,
indicating that there is a road of length z connecting house x and house
y.
The following M lines are the queries. Each line contains an
integer Q, asking that at most how many people can take part in Bob’s
race according to the above mentioned rules and under the condition that
the“race difference”is no more than Q.
The input ends with N = 0 and M = 0.
(N<=50000 M<=500 1<=x,y<=N 0<=z<=5000 Q<=10000000)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string.h>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<cmath>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const double PI=acos(-1.0);
const double eps=0.0000000001;
const int N=+;
int head[N];
int dp1[N][],dp2[N][];
int tot;
int n;
struct node{
int to,next,w;
}edge[N<<];
int dp[N][];
int vis[N];
int a[N];
void init(){
memset(head,-,sizeof(head));
memset(dp,,sizeof(dp));
tot=;
}
void add(int u,int v,int w){
edge[tot].to=v;
edge[tot].next=head[u];
edge[tot].w=w;
head[u]=tot++;
}
void DFS(int u,int fa){
int maxx1=;
int maxx2=;
for(int i=head[u];i!=-;i=edge[i].next){
int v=edge[i].to;
int w=edge[i].w;
if(v==fa)continue;
DFS(v,u);
int maxx=dp[v][]+w;
if(maxx>=maxx1){
maxx2=maxx1;
maxx1=maxx;
vis[u]=v;
}
else if(maxx>maxx2){
maxx2=maxx;
}
// cout<<u<<" "<<v<<" "<<maxx1<<" "<<maxx2<<endl;
}
dp[u][]=maxx1;
dp[u][]=maxx2;
}
void DFS1(int u,int fa){
for(int i=head[u];i!=-;i=edge[i].next){
int v=edge[i].to;
int w=edge[i].w;
if(v==fa)continue;
if(vis[u]==v){
dp[v][]=max(dp[u][]+w,dp[u][]+w);
}
else{
dp[v][]=max(dp[u][]+w,dp[u][]+w);
}
DFS1(v,u);
}
}
void init_RMQ(){
for(int i=;i<=n;i++){
dp1[i][]=a[i];
dp2[i][]=a[i];
}
for(int j=;(<<j)<=n;j++)
for(int i=;i+(<<j)-<=n;i++){
dp1[i][j]=max(dp1[i][j-],dp1[i+(<<(j-))][j-]);
dp2[i][j]=min(dp2[i][j-],dp2[i+(<<(j-))][j-]);
}
}
int getRMQ(int L,int R){
int k=;
while((<<(k+))<=R-L+)k++;
int maxx=max(dp1[L][k],dp1[R-(<<k)+][k]);
int minn=min(dp2[L][k],dp2[R-(<<k)+][k]);
return maxx-minn;
} int main(){
int m;
while(scanf("%d%d",&n,&m)!=EOF){ if(m==&&n==)break;
init();
for(int i=;i<=n;i++){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
DFS(,-);
DFS1(,-);
for(int i=;i<=n;i++){
a[i]=max(dp[i][],dp[i][]);
//cout<<a[i]<<" ";
}
init_RMQ();
for(int i=;i<m;i++){
int x;
scanf("%d",&x);
int l=;
int r=;
int ans=;
while(l<=n&&r<=n){
while(r<=n&&getRMQ(l,r)<=x){
r++;
}
ans=max(ans,r-l);
l++;
}
cout<<ans<<endl;
}
}
}
hdu 4123(树形dp+倍增)的更多相关文章
- hdu 4123 树形DP+RMQ
http://acm.hdu.edu.cn/showproblem.php? pid=4123 Problem Description Bob wants to hold a race to enco ...
- hdu 4123 树形DP+单调队列
http://acm.hust.edu.cn/vjudge/problem/25790 这题基本同poj 3162 要注意mx,mx2,vx,vx2每次都要初始化 #include <iostr ...
- 【bzoj2500】幸福的道路 树形dp+倍增RMQ+二分
原文地址:http://www.cnblogs.com/GXZlegend/p/6825389.html 题目描述 小T与小L终于决定走在一起,他们不想浪费在一起的每一分每一秒,所以他们决定每天早上一 ...
- HDU 1520 树形dp裸题
1.HDU 1520 Anniversary party 2.总结:第一道树形dp,有点纠结 题意:公司聚会,员工与直接上司不能同时来,求最大权值和 #include<iostream> ...
- HDU 1561 树形DP入门
The more, The Better Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Oth ...
- Codeforces 418d Big Problems for Organizers [树形dp][倍增lca]
题意: 给你一棵有n个节点的树,树的边权都是1. 有m次询问,每次询问输出树上所有节点离其较近结点距离的最大值. 思路: 1.首先是按照常规树形dp的思路维护一个子树节点中距离该点的最大值son_di ...
- HDU 2196树形DP(2个方向)
HDU 2196 [题目链接]HDU 2196 [题目类型]树形DP(2个方向) &题意: 题意是求树中每个点到所有叶子节点的距离的最大值是多少. &题解: 2次dfs,先把子树的最大 ...
- HDU 1520 树形DP入门
HDU 1520 [题目链接]HDU 1520 [题目类型]树形DP &题意: 某公司要举办一次晚会,但是为了使得晚会的气氛更加活跃,每个参加晚会的人都不希望在晚会中见到他的直接上司,现在已知 ...
- codevs 1380/HDU 1520 树形dp
1380 没有上司的舞会 时间限制: 1 s 空间限制: 128000 KB 题目等级 : 钻石 Diamond 题解 查看运行结果 回到问题 题目描述 Description Ural大学有N个职员 ...
- HDU 5834 [树形dp]
/* 题意:n个点组成的树,点和边都有权值,当第一次访问某个点的时候获得利益为点的权值 每次经过一条边,丢失利益为边的权值.问从第i个点出发,获得的利益最大是多少. 输入: 测试样例组数T n n个数 ...
随机推荐
- vue学习总结(简单介绍)
声明式渲染 Vue.js 的核心是一个允许采用简洁的模板语法来声明式地将数据渲染进 DOM 的系统: <div id="app"> {{ message }} < ...
- C语言编辑编译及集成开发环境
C语言编辑编译及集成开发环境 编辑器 在不同的操作系统上使用不同的编辑器,保存源代码文件时,文件名应指出程序的功能扩展名应为.c. 编译器 编译器把源代码编译成机器语言的二进制指令即目标代码生成目标文 ...
- Deepin系统关于每次启动终端都要输入source /etc/profile的问题
关于每次启动终端都要输入source /etc/profile的问题 当我在Deepin系统中下载了node以及npm之后,我为了将node导入到系统文件,使用了以下命令sudo gedit ``/e ...
- Opencv下双线性插值法进行图像放缩
关于图像放缩的算法有很多,本文主要介绍双线性插值法进行图像放缩,本文参考了: http://www.cnblogs.com/funny-world/p/3162003.html 我们设源图像src的大 ...
- 正确的在循环list的时候删除list里面的元素
s = [1,2,3,4,5] for i in s: s.remove(i) print(s) 输出结果:[2, 4] 1.当第一次删除后,后面的元素会前移,此时s=[2,3,4,5], 2.然 ...
- PAT 1142 Maximal Clique
A clique is a subset of vertices of an undirected graph such that every two distinct vertices in the ...
- 九九乘法表-Java
public class Test1 { public static void main(String[] args){ for(int i=1;i<=9;i++){ for(int j=1;j ...
- windows下通过navicat for mysql连接centos6.3-64bit下的MySQL数据库
一.centos下MySQL安装 按照命令依次安装以下文件: mysql-devel 开发用到的库以及包含文件 mysql mysql 客户端 mysql-server 数据库服务器 yum inst ...
- 清空所有Session
//清空所有Session request.getSession().invalidate();
- HDU 4451 容斥原理
题目大意: n件衣服,m条裤子,k双鞋子进行搭配 妈妈指明了哪些衣服和裤子不能搭配,哪些裤子和鞋子不能搭配,问最后有几种搭配方法 先假设都能搭配 n*m*k 每次遇到衣服和裤子不能搭的,就要减一次k, ...