HDU 6107 Typesetting (倍增)
Typesetting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 144 Accepted Submission(s): 72
The page width is fixed to W characters. In order to make the article look more beautiful, Yellowstar has made some rules:
1. The fixed width of the picture is pw. The distance from the left side of the page to the left side of the photo fixed to dw, in other words, the left margin is dw, and the right margin is W - pw - dw.
2. The photo and words can't overlap, but can exist in same line.
3. The relative order of words cannot be changed.
4. Individual words need to be placed in a line.
5. If two words are placed in a continuous position on the same line, then there is a space between them.
6. Minimize the number of rows occupied by the article according to the location and height of the image.
However, Yellowstar has not yet determined the location of the picture and the height of the picture, he would like to try Q different locations and different heights to get the best look. Yellowstar tries too many times, he wants to quickly know the number of rows each time, so he asked for your help. It should be noted that when a row contains characters or pictures, the line was considered to be occupied.
Each case begins with one line with four integers N, W, pw, dw : the number of words, page width, picture width and left margin.
The next line contains N integers ai, indicates i-th word consists of ai characters.
The third line contains one integer Q.
Then Q lines follow, each line contains the values of xi and hi, indicates the starting line and the image height of the image.
Limits
T≤10
1≤N,W,Q≤105
1≤pw,ai≤W
0≤dw≤W−pw
2 7 4 3
1 3
3
1 2
2 2
5 2
3 8 2 3
1 1 3
1
1 1
3
3
1
#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 1e5+;;
const int M = ;
const int mod = 1e9+;
const int mo=;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
int n,w,q,pw,dw;
int a[N],lto[N],rto[N];
int h[N],s[N][M],t[N][M];
void solve(int len,int to[]){
for(int i=,j=,x=-;i<=n;i++){
while(x+a[j]+<=len)x+=a[j++]+;
to[i]=j;x-=a[i]+;
}
}
int main() {
int T;
scanf("%d",&T);
while(T--){
scanf("%d%d%d%d",&n,&w,&pw,&dw);
rep(i,,n-)scanf("%d",&a[i]);
a[n]=w+;
solve(w,lto);
rep(i,,n)s[i][]=lto[i];
rep(j,,M-)rep(i,,n)s[i][j]=s[s[i][j-]][j-];
h[n]=;
for(int i=n-;i>=;i--)h[i]=h[lto[i]]+;
solve(dw,lto);solve(w-dw-pw,rto);
rep(i,,n)t[i][]=rto[lto[i]];
rep(j,,M-)rep(i,,n)t[i][j]=t[t[i][j-]][j-];
scanf("%d",&q);
while(q--){
int x,hh,res=;
scanf("%d%d",&x,&hh);
int ans=;
ans+=min(--x,h[]);
rep(j,,M-)if(x>>j & )res=s[res][j];
rep(j,,M-)if(hh>>j & )res=t[res][j];
ans+=hh+h[res];
printf("%d\n",ans);
}
}
return ;
}
HDU 6107 Typesetting (倍增)的更多相关文章
- HDU 6107 - Typesetting | 2017 Multi-University Training Contest 6
比赛的时候一直念叨链表怎么加速,比完赛吃饭路上突然想到倍增- - /* HDU 6107 - Typesetting [ 尺取法, 倍增 ] | 2017 Multi-University Train ...
- HDU 6107 Typesetting
Problem Description Yellowstar is writing an article that contains N words and 1 picture, and the i- ...
- Typesetting HDU - 6107
Yellowstar is writing an article that contains N words and 1 picture, and the i-th word contains aia ...
- hdu 5726 GCD 倍增+ 二分
题目链接 给n个数, 定义一个运算f[l,r] = gcd(al, al+1,....ar). 然后给你m个询问, 每次询问给出l, r. 求出f[l, r]的值以及有多少对l', r' 使得f[l, ...
- HDU 4822 Tri-war(LCA树上倍增)(2013 Asia Regional Changchun)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4822 Problem Description Three countries, Red, Yellow ...
- HDU 5875 Function 【倍增】 (2016 ACM/ICPC Asia Regional Dalian Online)
Function Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total ...
- hdu 2586 How far away ?倍增LCA
hdu 2586 How far away ?倍增LCA 题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=2586 思路: 针对询问次数多的时候,采取倍增 ...
- HDU - 6394 Tree(树分块+倍增)
http://acm.hdu.edu.cn/showproblem.php?pid=6394 题意 给出一棵树,然后每个节点有一个权值,代表这个点可以往上面跳多远,问最少需要多少次可以跳出这颗树 分析 ...
- hdu 6394 Tree (2018 Multi-University Training Contest 7 1009) (树分块+倍增)
链接: http://acm.hdu.edu.cn/showproblem.php?pid=6394 思路:用dfs序处理下树,在用分块,我们只需要维护当前这个点要跳出这个块需要的步数和他跳出这个块去 ...
随机推荐
- 关于ng-if的理论性知识你了解多少?
ng-if简介: ● 使用ng-if指令可以完全根据表达式的值在DOM中生成或移除一个元素.如果赋值给ng-if 的表达式的值是false,那对应的元素将会从DOM中移除,否则对应元素的一个克隆将被重 ...
- java分页通用篇
一.创建分页通用类 package com.dkyw.util; import java.util.List; public class Page<T> { private int tot ...
- 2.0 docker安装
问题列表: Error: Cannot retrieve metalink for repository: epel. Please verify its path and try again 解:处 ...
- Part2-HttpClient官方教程-Chapter7-高级主题(Advanced topics) (HTTP Caching)
原文链接 7.1 自定义客户端连接 在某些情况下,为了能够处理非标准的.不兼容的行为,可能需要自定义HTTP消息通过网络传输的方式,而不是使用HTTP参数.例如,对于web爬虫,可能有必要迫使Http ...
- linux驱动基础系列--Linux下Spi接口Wifi驱动分析
前言 本文纯粹的纸上谈兵,我并未在实际开发过程中遇到需要编写或调试这类驱动的时候,本文仅仅是根据源码分析后的记录!基于内核版本:2.6.35.6 .主要是想对spi接口的wifi驱动框架有一个整体的把 ...
- java===java基础学习(5)---文件读取,写入操作
文件的写入读取有很多方法,今天学到的是Scanner和PrintWriter 文件读取 Scanner in = new Scanner(Paths.get("file.txt") ...
- 自定义shell开头PS1
vim /etc/profile export PS1="flag:\W \u\$" \h是主机名,并不全,域 \W是当前所在目录名 \u 是当前shell用户名
- 2017多校第7场 HDU 6127 Hard challenge 极角排序,双指针
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6127 题意:平面直角坐标系上有n个整点,第i个点有一个点权val,坐标为(xi,yi),其中不存在任 ...
- C核心 那些个关键字
概述 - C语言老了 目前而言(2017年5月12日) C语言中有 32 + 5 + 7 = 44 个关键字. 具体如下 O(∩_∩)O哈哈~ -> C89关键字 char short int ...
- SVN--版本控制系统
引言 SVN是Subversion的简称,是一个开放源代码的版本控制系统,相较于RCS.CVS,它采用了分支管理系统,它的设计目标就是取代CVS.互联网上很多版本控制服务已从CVS迁移到Subvers ...