Codeforces 650B Image Preview
Vasya's telephone contains n photos. Photo number 1 is currently opened on the phone. It is allowed to move left and right to the adjacent photo by swiping finger over the screen. If you swipe left from the first photo, you reach photo n. Similarly, by swiping right from the last photo you reach photo 1. It takes a seconds to swipe from photo to adjacent.
For each photo it is known which orientation is intended for it — horizontal or vertical. Phone is in the vertical orientation and can't be rotated. It takes b second to change orientation of the photo.
Vasya has T seconds to watch photos. He want to watch as many photos as possible. If Vasya opens the photo for the first time, he spends 1 second to notice all details in it. If photo is in the wrong orientation, he spends b seconds on rotating it before watching it. If Vasya has already opened the photo, he just skips it (so he doesn't spend any time for watching it or for changing its orientation). It is not allowed to skip unseen photos.
Help Vasya find the maximum number of photos he is able to watch during T seconds.
The first line of the input contains 4 integers n, a, b, T (1 ≤ n ≤ 5·105, 1 ≤ a, b ≤ 1000, 1 ≤ T ≤ 109) — the number of photos, time to move from a photo to adjacent, time to change orientation of a photo and time Vasya can spend for watching photo.
Second line of the input contains a string of length n containing symbols 'w' and 'h'.
If the i-th position of a string contains 'w', then the photo i should be seen in the horizontal orientation.
If the i-th position of a string contains 'h', then the photo i should be seen in vertical orientation.
Output the only integer, the maximum number of photos Vasya is able to watch during those T seconds.
4 2 3 10
wwhw
2
5 2 4 13
hhwhh
4
5 2 4 1000
hhwhh
5
3 1 100 10
whw
0
In the first sample test you can rotate the first photo (3 seconds), watch the first photo (1 seconds), move left (2 second), rotate fourth photo (3 seconds), watch fourth photo (1 second). The whole process takes exactly 10 seconds.
Note that in the last sample test the time is not enough even to watch the first photo, also you can't skip it.
-------------------------------------------------------
仔细读题啊!
Solution:
看了题解过的,果然萎得不行。
我能想到的:
看的图片必然连续,因为每次只能选择看右边或左边还没看的那张图。
最后看的图的分布:
#(n-l)#(n-l+1)#(n-l+2)#...#(n-1)#(n)#(1)#(2)#(3)...#(r)
|<--------------------------------------------
|_____________________________------------------->
--------------------- >|
<-----------------------------------------————————|
(#表示图片,括号内是其ID,虚线表示看图,实线表示跳过)
但是我TM就是没看出来要达到最有解,翻图的路线只有两种情况:
对于给定的(l, r)看图的时间(包括调整方向与看)是固定,所以总时间取决于翻图的次数。
---------------------------------------------------------------------------------------------------------------------
所以最优解只有四种情况:
1.一直向右翻
2.一直向左翻
3.先向左翻,再向右翻
4.先向右翻再向左翻
后两种情况可以用双指针(two-pointers)做。
------------------------------------------------------------------
Implementation:
two-pointers写得磕磕绊绊,coding弱得不行,sigh.
(Ignore the comment like a compiler :D)
#include <bits/stdc++.h>
using namespace std; typedef long long LL; const int N(5e5+); char o[N];
int n, a, b, T; int cost(int x){
return +(o[x]=='w')*b+(bool)x*a;
} int main(){
for(;cin>>n>>a>>b>>T>>o;){
int ans=;
for(;cost()<=T;){
int t=T-cost(), i, j;
//two-pointers
for(i=; i<n; i++){
if(cost(i)>t) break;
t-=cost(i);
}
ans=max(ans, i);
if(ans==n) break; // L.I.:
// i-1: current position to the righ;
for(t-=(i-)*a, j=; i>=; t+=cost(i-)+a, i--){
// L.I.:
// j:offset to the left
// n-j-1:current left position
for(;j<n-i && cost(n-j-)<=t; j++){
t-=cost(n-j-);
}
ans=max(ans, i+j);
} if(ans==n) break; t=T-cost();
for(i=; i<n; i++){
if(cost(n-i)>t) break;
t-=cost(n-i);
} ans=max(ans, i);
if(ans==n) break; // n-i+1: last position
for(t-=(i-)*a, j=; i>=; t+=cost(n-i+)+a, i--){//one step back
for( ; j<n-i && cost(j+)<=t; j++){
t-=cost(j+);
}
ans=max(ans, i+j);
}
break;
} cout<<ans<<'\n';
}
return ;
}
Codeforces 650B Image Preview的更多相关文章
- codeforces 650B . Image Preview 二分
题目链接 B. Image Preview time limit per test 1 second memory limit per test 256 megabytes input standar ...
- Codeforces 650B Image Preview(尺取法)
题目大概说手机有n张照片.通过左滑或者右滑循环切换照片,滑动需要花费a时间:看一张照片要1时间,而看过的可以马上跳过不用花时间,没看过的不能跳过:有些照片要横着看,要花b时间旋转方向.那么问T时间下最 ...
- 汕头市队赛 C KMP codeforces B. Image Preview
汕头市队赛题目传送门 codeforces题目传送门 这道题我的做法是 尝试先往左走然后往右走 或者先往右走然后往左走 然后注意一下枚举顺序就okay啦 #include<cstdio> ...
- Codeforces 651D Image Preview【二分+枚举】
题意: 若干张照片,从头开始可以向左右两边读,已经读过的不需要再读,有的照片需要翻转,给定读.滑动和翻转消耗的时间,求在给定时间内最多能读多少页? 分析: 首先明确,只横跨一次,即先一直读一边然后再一 ...
- Codeforces Round #345 (Div. 2) D. Image Preview 暴力 二分
D. Image Preview 题目连接: http://www.codeforces.com/contest/651/problem/D Description Vasya's telephone ...
- codeforces 650D D. Image Preview (暴力+二分+dp)
题目链接: http://codeforces.com/contest/651/problem/D D. Image Preview time limit per test 1 second memo ...
- Codeforces Round #345 D. Image Preview(二分)
题目链接 题意:看一个图片需要1单位时间,如果是 w 需要翻转 b 时间,切换到相邻位置(往左或者往右)需要 a 时间,求T时间最多能看几张图片 从第一个开始向右走看若干个图片然后往如果往左走就不会再 ...
- Codeforces Round #345 (Div. 1) B. Image Preview
Vasya's telephone contains n photos. Photo number 1 is currently opened on the phone. It is allowed ...
- CodeForces - 651D:Image Preview (双指针&)
Vasya's telephone contains n photos. Photo number 1 is currently opened on the phone. It is allowed ...
随机推荐
- C/C++ 常用工具集
1. c++filt //注意:就是这个名字 "c++file". 能把c++的函数签名转换成代码形参格式: 如:# c++filt _ZNSt4priv17_Rb_tree_i ...
- svn命令行修改已提交的版本备注
svn命令行修改已提交的版本备注 参考文章: stackoverflow.com/questions/304383/how-do-i-edit-a-log-message-that-i-already ...
- javarebel热部署 (转)
Java web开发部署效率浅析 在进行java web程序开发过程中,经常遇到这种问题,修改一个java文件(*.java),需要重启web服务器(如tomcat,weblogic等),部署项目.而 ...
- LUA __call
1. ev={} . functin ev.__call() . print "called from ev" . end . . setmetatable(ev, ev) . . ...
- Android使用service后台更新计划任务
Service是Android的四大组件之一,这里就不再过多的去描述,下面主要实现启动应用时候利用service后台执行计划任务,退出应用后,关闭service,只存在整个应用的周期中. 首先使用se ...
- struct2cell
函数功能:把结构体转换为元胞数组. 语法格式: c = struct2cell(s) 如果s是m*n(m行n列)的二维的结构体数组,每个结构体含有p个域,则转换得到一个p*m*n的元胞数组c. 如果s ...
- 20135223/20135234/20135229小组——亚博 Arduino智能小车实践报告
实验名称:Arduino智能小车组装和综合测试 实验小组成员:20135223何伟钦 20135234马启扬 20135229吕松鸿 实验日期:2015.10.27—2015.11.3 实验时长:24 ...
- 20145215《Java程序设计》第3周学习总结
20145215<Java程序设计>第三周学习总结 教材学习内容总结 认识对象 类类型 在学习第三章的时候,我们知道Java可区分为基本类型和类类型两大类型系统,其中类类型也称为参考类型. ...
- C# Winform实现炫酷的透明动画界面
做过.NET Winform窗体美化的人应该都很熟悉UpdateLayeredWindow吧,UpdateLayeredWindow可以实现窗体的任意透明,效果很好,不会有毛边.不过使用这个API之后 ...
- ionic 项目笔记
最近公司在用ionic 做 微站,项目中难免遇到一些问题.总结如下: 1. 改了Slidebox 动态绑定图片时,页面会显示一片空白,改变窗口大小的时候,图片就出来了,说明动态绑定图片时, ...