"BZOJ4385" [Poi2015]wilcze do?y "monotone queue" "prefix and" "" "and" "".

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Exercises

If the interval [j, I] is fixed, then it must be that the largest part of the weight becomes 0.

Using a monotone queue to maintain a sub-Piete with the maximal weights within a period (here the right endpoint subscript, set to X), enumerate right end I, use the ruler method to calculate J.

An interval [j, I] is valid for the condition of Sum[i]-sum[j-1]-(sum[x]-sum[x-d]) <= p.

Complexity of:

Time complexity: Because each point traverses up to 2 times, the complexity is O (n).

Space complexity: O (N)

RE:

I forgot to write H <= t when I updated J. If you write the right shot, you should be able to shoot it.

GET:

If d = 0, then it is the most bare ruler.

From the ruler method, think of the following methods, but can prove to be wrong:

1 First find the longest interval, meet the interval and not less than P, and then the maximum weight of the section into 0, and then expand.

counter example: n = 9,p = 3,d = 1,10 1 1 1 10 1 2 1 1, the answer should be 4.

2 Find the interval with the maximum weight of D, change to 0, and then expand from here.

counter example: n = 8,p = 19,d = 2,10 11 12 10 1 1 1 1, the answer should be 7.

So it is not possible to maintain the static, consider adding a number, and then think of the standard solution, but do not know how to maintain a fixed length of the interval of the maximum value of the sub-segment.

You should use the monotone queue + prefix and.

Again by the monotone queue + prefix and stuck, it seems to be summed up.

/* telekinetic Forest Guard */#include <cstdio> #include <cstring> #include <algorithm>using namespace Std;typedef Long Long ll;const int maxn = 2000005;int N, D; LL p, sum[maxn];int q[maxn];template <class numtype>inline void Read (Numtype &x) {bool F = 0; x = 0; char ch = g Etchar (); for (Ch < ' 0 ' | | ch > ' 9 '; ch = getchar ()) F = ch = = '-'? 1:0;for (; Ch >= ' 0 ' && ch <= ' 9 '; ch = getchar ()) x = x * + ch-' 0 '; if (f) x =-X;} int main () {read (n); Read (P); Read (d); for (int i = 1; I <= n; i++) read (Sum[i]), sum[i] + = sum[i-1];int h = 1, t = 0, ans = 0;for (int i = d, j = 1; I <= n; i++) {for (; h <= t && sum[i]-sum[i-d] > Sum[q[t]]-sum[q[t]- D];  t--) Q[++t] = I;for (; h <= t && sum[i]-sum[j-1]-(sum[q[h]]-sum[q[h]-d]) > P; j + +) if (Q[h]-D <= j) H++;ans = max (ans, i-j + 1);} printf ("%d\n", ans); return 0;}

"BZOJ4385" [Poi2015]wilcze do?y "monotone queue" "prefix and" "" "and" "".