CodeForces

D. Winter Is Coming

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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 62 -5 1 3 0 0 -4 -3 1 0

output

3

题意： 冬天到了，现在知道n天的温度，现在有夏天轮胎和冬天轮胎，冬天轮胎只能用k天。

已知，当温度<0时要用冬天轮胎，当温度>=0时可以用冬天轮胎也可以用夏天轮胎，问你最少需要换轮胎的次数。

并且最初车上装的是夏天轮胎。如果可能输出最小换胎次数，不然输出-1.

思路：用sum记录温度低于0的天数，用res记录sum-k，意思就是我们有res天可以在温度>=0的时候用冬天的轮胎，这样就避免总是换轮胎，

而很容易可以看出来如果温度>=0的时候不用冬胎的话，也就是最坏情况 2*sum。但是如果我们把相邻的两个（这里相邻中间可以有温度》=0的时候）温度<0的

天数连在一起不换胎那么就可以减少2次换胎的次数。那么这两个<0的时间中的>=0的天数也会从res中减去，所以为了使换胎次数最少，我们贪心的依据就是

两个相邻的<0的天数之间的>=0的天数越少越好。

但是最后还有一个特殊情况就是如果最后用的冬天的轮胎，并且res>=最后一个<0之后的>=0的天数，那么最后就可以不用换轮胎了，直接用到最后，这样最终的

结果就可以-1  (sum-1)。

代码：

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define N 200005#define inf 0x3f3f3fusing namespace std;struct node{int left,right;int c;}dong[N];int a[N];int n,k,cnt;bool cmp(node a,node b){return a.c<b.c;}int main(){int i,j;int x,left,right;scanf("%d %d",&n,&k);int sum=0;for(i=1;i<=n;i++){scanf("%d",&x);if(x<0){a[i]=-1;sum++;}else a[i]=0;}cnt=0;//printf("sum: %d\n",sum);left=right=-1;for(i=1;i<=n;i++){if(a[i]<0){left=i;break;}}i++;int count=0;for(;i<=n;i++){if(a[i]<0){right=i;dong[++cnt].left=left;dong[cnt].right=right;dong[cnt].c=count;count=0;left=right;}else count++;}/*printf("cnt: %d\n",cnt);for(i=1;i<=cnt;i++) printf("%d %d %d\n",dong[i].left,dong[i].right,dong[i].c);printf("\n");*/int en=n-left;//if(cnt!=0) en=n-dong[cnt].right;sort(dong+1,dong+cnt+1,cmp);if(k>=sum){int res=k-sum;sum*=2;//printf("sum: %d\n",sum);for(i=1;i<=cnt;i++){if(dong[i].c<=res){sum-=2;res-=dong[i].c;}}if(res>=en){sum--;}printf("%d\n",sum);}else printf("-1\n");return 0;}