【23.26%】【codeforces 747D】Winter Is Coming
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
The winter in Berland lasts n days. For each day we know the forecast for the average air temperature that day.
Vasya has a new set of winter tires which allows him to drive safely no more than k days at any average air temperature. After k days of using it (regardless of the temperature of these days) the set of winter tires wears down and cannot be used more. It is not necessary that these k days form a continuous segment of days.
Before the first winter day Vasya still uses summer tires. It is possible to drive safely on summer tires any number of days when the average air temperature is non-negative. It is impossible to drive on summer tires at days when the average air temperature is negative.
Vasya can change summer tires to winter tires and vice versa at the beginning of any day.
Find the minimum number of times Vasya needs to change summer tires to winter tires and vice versa to drive safely during the winter. At the end of the winter the car can be with any set of tires.
Input
The first line contains two positive integers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — the number of winter days and the number of days winter tires can be used. It is allowed to drive on winter tires at any temperature, but no more than k days in total.
The second line contains a sequence of n integers t1, t2, …, tn ( - 20 ≤ ti ≤ 20) — the average air temperature in the i-th winter day.
Output
Print the minimum number of times Vasya has to change summer tires to winter tires and vice versa to drive safely during all winter. If it is impossible, print -1.
Examples
input
4 3
-5 20 -3 0
output
2
input
4 2
-5 20 -3 0
output
4
input
10 6
2 -5 1 3 0 0 -4 -3 1 0
output
3
Note
In the first example before the first winter day Vasya should change summer tires to winter tires, use it for three days, and then change winter tires to summer tires because he can drive safely with the winter tires for just three days. Thus, the total number of tires’ changes equals two.
In the second example before the first winter day Vasya should change summer tires to winter tires, and then after the first winter day change winter tires to summer tires. After the second day it is necessary to change summer tires to winter tires again, and after the third day it is necessary to change winter tires to summer tires. Thus, the total number of tires’ changes equals four.
【题目链接】:http://codeforces.com/contest/747/problem/D
【题解】
找到最左边的l,满足t[l]<0,找到最右边的r,满足t[r]<0;
然后在l..r这个区间里面
找连续的t>=0的块
把每个块记录起来;
然后在l..r里面
统计t>=0的个数->a
统计t<0的个数->b
如果b>k则无解;
如果b<=k;
设rest = k-b;
则看看l..r里面哪一些>=0的连续块能够用rest抵消掉;
即让雪地胎在温度大于等于0的时候用.这样这一块就不用换胎了(减少了2次换胎);
显然我们是要让块数最多。
那么每次找块的长度最短的减就好了;
减掉后标记这一块被减掉了;否则
对那些没被减掉的块,答案都递增2;
还有在l之前肯定是夏天胎。则肯定要递增一次;
如果在r后面还有温度大于0的,则要看能不能直接走到最后->即rest还有没有剩余;否则还要递增答案1;
这个可以把r+1..n单独作为一块加到已经升序排完后的>=0块vecotr后面;因为这个对答案的递减作用只有1;而哪些>=0的块对答案的递减作用有2;所以最后再考虑它.
全都是正数的情况特判下.
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int MAXN = 2e5+100;
const int INF = 21e8;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
struct abc
{
int l,r,len,flag;
};
int n,k,t[MAXN],trick[MAXN];
int cnt = 0;
vector <abc> a;
bool cmp(abc a,abc b)
{
return a.len<b.len;
}
int main()
{
//freopen("F:\\rush.txt","r",stdin);
rei(n);rei(k);
rep1(i,1,n)
rei(t[i]);
int l = -1,r = -1;
rep1(i,1,n)
if (t[i]<0)
{
l = i;
break;
}
rep2(i,n,1)
if (t[i] <0)
{
r = i;
break;
}
if (l==-1)
{
cout << 0<<endl;
return 0;
}
rep1(i,l,r)
{
if (t[i]>=0)
{
int ll = i,rr = i;
while (rr+1<=r && t[rr+1]>=0) rr++;
a.pb({ll,rr,rr-ll+1,1});
i = rr;
}
}
sort(a.begin(),a.end(),cmp);
if (r<n)
a.pb({r+1,n,n-(r+1)+1,1});
else
a.pb({r+1,n,INF,0});
int aa = 0,bb = 0;
rep1(i,l,r)
if (t[i] >=0)
aa++;
else
bb++;
int now = k-bb;
if (now <0)
{
puts("-1");
return 0;
}
int len = a.size();
rep1(i,0,len-1)
{
if (now >= a[i].len)
{
now-=a[i].len;
a[i].flag = 0;
}
}
int ans = 0;
rep1(i,0,len-2)
if (a[i].flag==1)
ans +=2;
if (a[len-1].flag==1)
ans += 1;
ans+=1;
cout << ans << endl;
return 0;
}
【23.26%】【codeforces 747D】Winter Is Coming的更多相关文章
- 【23. 合并K个排序链表】【困难】【优先队列/堆排序】
合并 k 个排序链表,返回合并后的排序链表.请分析和描述算法的复杂度. 示例: 输入: [ 1->4->5, 1->3->4, 2->6] 输出: 1->1-> ...
- 【 BowWow and the Timetable CodeForces - 1204A 】【思维】
题目链接 可以发现 十进制4 对应 二进制100 十进制16 对应 二进制10000 十进制64 对应 二进制1000000 可以发现每多两个零,4的次幂就增加1. 用string读入题目给定的二进制 ...
- Codeforces 747D:Winter Is Coming(贪心)
http://codeforces.com/problemset/problem/747/D 题意:有n天,k次使用冬天轮胎的机会,无限次使用夏天轮胎的机会,如果t<=0必须使用冬轮,其他随意. ...
- 【26.83%】【Codeforces Round #380C】Road to Cinema
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- 【20.23%】【codeforces 740A】Alyona and copybooks
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- 【23.33%】【codeforces 557B】Pasha and Tea
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- 【23.39%】【codeforces 558C】Amr and Chemistry
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- 【30.23%】【codeforces 552C】Vanya and Scales
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
- 【26.09%】【codeforces 579C】A Problem about Polyline
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
随机推荐
- LintCode刷题笔记-- Update Bits
标签: 位运算 描述: Given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to set ...
- ELK学习之jdk和jre的区别
JRE: Java Runtime EnvironmentJDK:Java Development KitJRE顾名思义是java运行时环境,包含了java虚拟机,java基础类库.是使用java语言 ...
- Spring MVC 关于controller的字符编码问题
在使用springMVC框架构建web应用,客户端常会请求字符串.整型.json等格式的数据,通常使用@ResponseBody注解使 controller回应相应的数据而不是去渲染某个页面.如果请求 ...
- QT自定义窗口
qt 中允许自定义窗口控件,使之满足特殊要求, (1)可以修改其显示,自行绘制 (2)可以动态显示 (3)可以添加事件,支持鼠标和键盘操作 自定义控件可以直接在QtDesigner里使用,可以直接加到 ...
- laravel 中使用tinker 验证驱动加载是否成功
在验证laravel 中 InvalidArgumentException Driver [WeiBo] not supported. public function weibo() { retu ...
- PHP实现购物车的思路和源码分析
正文内容 这里主要是记录下自己的购物车的思路,具体功能实现,但是尚未在实际项目中用到,不对之处欢迎指正 项目中需要添加购物车. 目录说明 buy.php 点击购买之后的操作 car.php 购物车,显 ...
- oracle总是使用索引的第一个列
如果索引是建立在多个列上, 只有在它的第一个列(leading column)被where子句引用时,优化器才会选择使用该索引. 译者按: 这也是一条简单而重要的规则. 见以下实例. SQL> ...
- 如何学习Python的一些总结
https://mp.weixin.qq.com/s/w0NoDiYfvtTy8N3BVoIVpw 为什么选择Python 经常会有同学问我为什么选择Python.我很喜欢这门语言,因为它的简洁灵活, ...
- mybatis 嵌套查询与懒加载
懒加载:对于页面有很多静态资源的情况下(比如网商购物页面),为了节省用户流量和提高页面性能,可以在用户浏览到当前资源的时候,再对资源进行请求和加载. fetchType="lazy" ...
- Math.abs( x )
Math.abs( x ) 下面是参数的详细信息: x : 一个数字 返回值: 返回一个数字的绝对值 <html> <head> <title>JavaScript ...