【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;
}