BZOJ2086: [Poi2010]Blocks

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=2086

如果一段前缀和>=k*len那么就是可行的。

于是对所有的a[i]-=k，求前缀和。然后从1~n维护一个单调递减的栈。

从n~1扫一遍，如果sum[i]-sum[q[top-1]]>=0，那么以q[top-1]作为左端点是更优的

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define inf int(1e9)
#define maxn 1005000
#define mm 30031
#define ll long long
using namespace std;
int n,m,top;
ll k;
ll a[maxn],sum[maxn];
int q[maxn];
ll read(){
    ll x=0,f=1; char ch=getchar();
    while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while (isdigit(ch)) {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
int main(){
    n=read(); m=read();
    rep(i,1,n) a[i]=read();
    rep(i,1,m){
        k=read();
        rep(j,1,n) sum[j]=sum[j-1]+a[j]-k;
        top=0;
        rep(j,1,n) if (sum[q[top]]>sum[j]) q[++top]=j;
        int ans=0;
        down(j,n,1){
            while (top&&sum[j]-sum[q[top-1]]>=0) top--;
            ans=max(ans,j-q[top]);
        } 
        if (i!=m) printf("%d ",ans); else printf("%d\n",ans);
    }
    return 0;
}