bzoj2216

决策单调性+整体二分

这里就是j<k且kj劣于j，j不会再选，所以我们整体二分

pos是因为从L->R中这个是最优点，所以对于mid+1->r选pos之前肯定不优，l->mid-1不会选>pos，因为每个位置都小于mid，并且pos->mid-1这段区间的决策点没有pos优，因为当前f[i]的i小于mid，选的决策的位置大于pos，由于i小于mid，所以sqrt(i-j)，j越大，下降越快，所以pos+1->mid-1肯定没pos优

#include<bits/stdc++.h>
using namespace std;
const int N = 500010;
int n;
int a[N], id[N];
double f1[N], f2[N];
double calc(int i, int j)
{
    return (double)a[j] - (double)a[i] + sqrt(abs((double)i - (double)j));
}
void solve(int l, int r, int L, int R, double *f)
{
    if(l > r) return;
    int mid = (l + r) >> 1, lim = min(mid, R), pos = lim;
    double mx = 0;
    for(int i = L; i <= lim; ++i) if(calc(mid, i) > mx) 
    {
        mx = calc(mid, i);
        pos = i;
    }
    f[id[mid]] = mx;
    solve(l, mid - 1, L, pos, f);
    solve(mid + 1, r, pos, R, f);
}
int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)    
    {
        id[i] = i;
        scanf("%d", &a[i]);        
    }
    solve(1, n, 1, n, f1);
    reverse(a + 1, a + n + 1);
    reverse(id + 1, id + n + 1);
    solve(1, n, 1, n, f2);
    for(int i = 1; i <= n; ++i) printf("%d\n", (int)ceil(max(f1[i], f2[i])));
    return 0;
}