D. 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.
题意：有两种轮胎，一种温度零下时和零上温度都能用，一种只能零上温度用，输入路程长度，输入有n个零下时用的轮胎个数。接下来是每段路的温度，问至少要更换多少次轮胎才能到达目的地。
题解：贪心即可，尽量少换轮胎，如果都是温度零上，就不需要更换轮胎。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int n,k,a[200008],i;
while(scanf("%d%d",&n,&k)!=EOF)
{
int vis[200008];
memset(vis,0,sizeof(vis));
int flag=0;
a[0]=1;
int change=0;
for(i=1; i<=n; i++)
{
scanf("%d",&a[i]);
if(a[i]<0)
{
flag++;
}
if((a[i-1]>=0&&a[i]<0)||(a[i-1]<0&&a[i]>=0))
{
change++;
}
}
if(flag==0)
{
printf("0
");
continue;
}
if(flag>k)
{
printf("-1
");
continue;
}
int j=0;
for(i=1; i<=n; i++)
{
vis[j]=a[i]>=0?vis[j]+1:vis[j]-1;
if((a[i]<0&&a[i+1]>=0)||(a[i]>=0&&a[i+1]<0))
{
j++;
if(i==n)j--;
}
}
// for(i=0;i<=j;i++)
// {
// printf("%d %d
",vis[i],i);
// }
if(j>1)
{
sort(vis+1,vis+j-1);
}
k=k-flag;
for(i=1;i<j;i++)
{
if(vis[i]>0&&vis[i]<=k)
{
change-=2;
k-=vis[i];
}
}
if(vis[j]<=k&&vis[j]>0)
{
change--;
k=k-vis[j];
}
// if((a[i-1]>=0&&a[i]<0)||(a[i-1]<0&&a[i]>=0))
// {
// change++;
// }
printf("%d
",change);
}
}