hdu 4747 Mex (2013 ACM/ICPC Asia Regional Hangzhou Online)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4747
思路:
比赛打得太菜了,不想写。。。。线段树莽一下
实现代码:
#include<iostream>
#include<cstdio>
#include<map>
#include<cmath>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid ll m = (l + r) >> 1
#define ll long long
const ll M = 3e5+;
ll sum[M<<],maxx[M<<],lazy[M<<],a[M],mex[M],n,Next[M];
map<ll,ll>mp;
void pushup(ll rt){
sum[rt] = sum[rt<<] + sum[rt<<|];
maxx[rt] = max(maxx[rt<<] , maxx[rt<<|]);
} void pushdown(ll rt,ll m){
if(lazy[rt]!=-){
lazy[rt<<] = lazy[rt];
lazy[rt<<|] = lazy[rt];
sum[rt<<] = lazy[rt]*(m-(m>>));
sum[rt<<|] = lazy[rt]*(m>>);
maxx[rt<<] = lazy[rt];
maxx[rt<<|] = lazy[rt];
lazy[rt] = -;
}
} void build(ll l,ll r,ll rt){
sum[rt] =; maxx[rt] = ;
lazy[rt] = -;
if(l == r){
sum[rt] = mex[l];
maxx[rt] = mex[l];
return ;
}
mid;
build(lson); build(rson);
pushup(rt);
} /*void ct(ll l,ll r,ll rt){
if(l == r){
cout << maxx[rt]<<" ";
return ;
}
mid;
ct(lson);
ct(rson);
}*/ void update(ll L,ll R,ll c,ll l,ll r,ll rt){
if(L <= l&&R >= r){
lazy[rt] = c;
maxx[rt] = c;
sum[rt] = c*(r-l+);
return ;
}
pushdown(rt,r-l+);
mid;
//cout<<l<<" "<<r<<" "<<sum[1]<<endl;
if(L <= m) update(L,R,c,lson);
if(R > m) update(L,R,c,rson);
pushup(rt);
} ll query(ll p,ll l,ll r,ll rt){
if(l == r) return l;
pushdown(rt,r-l+);
mid;
//cout<<maxx[rt<<1]<<" "<<p<<endl;
if(maxx[rt<<] > p) return query(p,lson);
else return query(p,rson);
} int main()
{
while(scanf("%I64d",&n)&&n){
ll tmp = ;
mp.clear();
for(ll i = ;i <= n;i ++){
scanf("%I64d",&a[i]);
mp[a[i]] = ;
while(mp.find(tmp) != mp.end()) tmp++;
mex[i] = tmp;
}
mp.clear();
for(ll i = n;i >= ;i --){
if(mp.find(a[i]) == mp.end()) Next[i] = n + ;
else Next[i] = mp[a[i]];
mp[a[i]] = i;
}
build(,n,);
ll ans = ;
for(ll i = ;i <= n;i ++){
ans += sum[];
if(maxx[] > a[i]){
ll l = query(a[i],,n,);
ll r = Next[i];
//cout<<l<<" "<<r<<endl;
if(l < r)
update(l,r-,a[i],,n,);
}
update(i,i,,,n,);
}
printf("%I64d\n",ans);
}
return ;
}
hdu 4747 Mex (2013 ACM/ICPC Asia Regional Hangzhou Online)的更多相关文章
- HDU 4747 Mex(线段树)(2013 ACM/ICPC Asia Regional Hangzhou Online)
Problem Description Mex is a function on a set of integers, which is universally used for impartial ...
- HDU 4745 Two Rabbits(最长回文子序列)(2013 ACM/ICPC Asia Regional Hangzhou Online)
Description Long long ago, there lived two rabbits Tom and Jerry in the forest. On a sunny afternoon ...
- HDU 4744 Starloop System(最小费用最大流)(2013 ACM/ICPC Asia Regional Hangzhou Online)
Description At the end of the 200013 th year of the Galaxy era, the war between Carbon-based lives a ...
- (并查集)Travel -- hdu -- 5441(2015 ACM/ICPC Asia Regional Changchun Online )
http://acm.hdu.edu.cn/showproblem.php?pid=5441 Travel Time Limit: 1500/1000 MS (Java/Others) Memo ...
- (二叉树)Elven Postman -- HDU -- 54444(2015 ACM/ICPC Asia Regional Changchun Online)
http://acm.hdu.edu.cn/showproblem.php?pid=5444 Elven Postman Time Limit: 1500/1000 MS (Java/Others) ...
- [2013 ACM/ICPC Asia Regional Hangzhou Online J/1010]hdu 4747 Mex (线段树)
题意: + ;];;;], seg[rt << | ]);)) * fa.setv;) * fa.setv;;], seg[rt << | ], r - l + );;, ...
- HDU 4757 Tree(可持久化字典树)(2013 ACM/ICPC Asia Regional Nanjing Online)
Problem Description Zero and One are good friends who always have fun with each other. This time, ...
- HDU 4729 An Easy Problem for Elfness(主席树)(2013 ACM/ICPC Asia Regional Chengdu Online)
Problem Description Pfctgeorge is totally a tall rich and handsome guy. He plans to build a huge wat ...
- HDU 4735 Little Wish~ lyrical step~(DLX搜索)(2013 ACM/ICPC Asia Regional Chengdu Online)
Description N children are living in a tree with exactly N nodes, on each node there lies either a b ...
随机推荐
- 20155318 《网络攻防》 Exp9 Web基础
20155318 <网络攻防> Exp9 Web基础 基础问题 SQL注入攻击原理,如何防御 就是通过把SQL命令插入到"Web表单递交"或"输入域名&quo ...
- # RocEDU.课程设计2018 第三周进展 博客补交
RocEDU.课程设计2018 第三周进展 博客补交 本周计划完成的任务 (1).本周计划完成在平板电脑上实现程序的功能,跟第二周计划完成任务基本相似. 本周实际完成情况 (1).实际完成情况还差最后 ...
- 【第三方插件】使用Topshelf创建Windows服务
概述 Topshelf是创建Windows服务的另一种方法,老外的一篇文章Create a .NET Windows Service in 5 steps with Topshelf通过5个步骤详细的 ...
- MiZ702学习笔记8——让MiZ702变身PC的方法
首先你需要一个安装好的linux系统,这里我用的是Ubuntu的虚拟机.VMWare的话,选择较高版本的成功率会高些(当然根据自己电脑的配置进行选择). 打开Ubuntu的虚拟机,找到一个叫做Disk ...
- matplotlib 雷达图2
说明 搞了一个最新版本的雷达图,比以前那个美观. 不多说,代码奉上: 完整代码 ''' matplotlib雷达图 ''' import numpy as np import matplotlib.p ...
- 为你的机器学习模型创建API服务
1. 什么是API 当调包侠们训练好一个模型后,下一步要做的就是与业务开发组同学们进行代码对接,以便这些‘AI大脑’们可以顺利的被使用.然而往往要面临不同编程语言的挑战,例如很常见的是调包侠们用Pyt ...
- stl源码剖析 详细学习笔记 算法(4)
//---------------------------15/03/31---------------------------- //lower_bound(要求有序) template<cl ...
- setBit testBit权限管理
1.jdk7文档解释 public boolean testBit(int n) Returns true if and only if the designated bit is set. (Com ...
- HTML 列表实例
41.无序列表本例演示无序列表.<h4>一个无序列表</h4><ul> <li>咖啡</li> <li>茶</li> ...
- tensorflow 曲线拟合
tensorflow 曲线拟合 Python代码: import numpy as np import tensorflow as tf import matplotlib.pyplot as plt ...