[BZOJ] 2442: [Usaco2011 Open]修剪草坪

\(ans=\sum\limits_{i=1}^n E[i] - mincost\)
因此我们求mincost即可

设f[i]表示前i个牛，第i头牛一定不在，满足条件的前提下的mincost
转移时一个区间最小值形式，下标单调递增，单调队列维护一下即可

答案就减去f[n+1]好了，这样保证1~n都是满足条件的了

 
#include<iostream>
#include<cstdio>
 
using namespace std;
 
 
inline bool isd(char c){return c>='0'&&c<='9';}
inline int rd(){
    int ret=0,f=1;char c;
    while(c=getchar(),!isd(c))f=c=='-'?-1:1;
    while(isd(c))ret=ret*10+c-'0',c=getchar();
    return ret*f;
}
 
const int MAXN = 100005;
typedef long long ll;
 
int n,m;
int a[MAXN];
ll sum,f[MAXN];
int q[MAXN],h,t;
int main(){
    n=rd();m=rd();
    for(int i=1;i<=n;i++)a[i]=rd(),sum+=a[i];
    m++;
    for(int i=1;i<=n+1;i++){
        while(h<=t&&q[h]<i-m)h++;
        f[i]=f[q[h]]+1ll*a[i];
        while(h<=t&&f[q[t]]>=f[i])t--;
        q[++t]=i;
    }
    cout<<sum-f[n+1];
}