【蓝桥杯 2022】最优清零方案

Posted by WHZ0325 on 2023-04-02, Viewed times

题目描述

给定一个长度为 $n$ 的序列 ${a_i}$ 和一个整数 $k$,一次操作可以将连续 $k$ 个整数减一或者将一个整数减一,求将所有整数变为 $0$ 的最少操作次数。

$1\le k\le n\le 10^6$,$0\le a_i\le 10^6$。

算法分析

啊,这题,我咋不会啊……

贪心策略是从左到右尽可能减连续 $k$ 个整数,操作次数和剩下零散的数加起来。

为啥是对的?如果我们要减连续 $k$ 个整数当然是越靠前越好,这样后面说不定可以再减一次,因为无论在哪里减对总和的影响都是相同的,如果在前面减区间的 $min$ 值更小,也不会妨碍后面有重叠的区间将少减的这几次补上。

代码实现

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import java.io.*;
public class Main {
static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
static BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
static int[] a, tg, min; static long[] sum;
static void build(int o, int l, int r) {
int mid = (l + r) >> 1;
if(l == r) { tg[o] = 0; sum[o] = min[o] = a[mid]; }
else {
build(o << 1, l, mid); build(o << 1 | 1, mid + 1, r);
min[o] = Math.min(min[o << 1], min[o << 1 | 1]);
sum[o] = sum[o << 1] + sum[o << 1 | 1];
}
}
static int ql, qr;
static int getMin(int o, int l, int r) {
if(ql <= l && r <= qr) return min[o];
int mid = (l + r) >> 1;
if(tg[o] != 0) {
tg[o << 1] += tg[o];
tg[o << 1 | 1] += tg[o];
min[o << 1] += tg[o];
min[o << 1 | 1] += tg[o];
sum[o << 1] += (long)tg[o] * (mid - l + 1);
sum[o << 1 | 1] += (long)tg[o] * (r - mid);
tg[o] = 0;
}
int ans = Integer.MAX_VALUE;
if(ql <= mid) ans = Math.min(ans, getMin(o << 1, l, mid));
if(mid + 1 <= qr) ans = Math.min(ans, getMin(o << 1 | 1, mid + 1, r));
return ans;
}
static int x;
static void modify(int o, int l, int r) {
if(ql <= l && r <= qr) {
tg[o] += x;
min[o] += x;
sum[o] += (long)x * (r - l + 1);
}
else {
int mid = (l + r) >> 1;
if(tg[o] != 0) {
tg[o << 1] += tg[o];
tg[o << 1 | 1] += tg[o];
min[o << 1] += tg[o];
min[o << 1 | 1] += tg[o];
sum[o << 1] += (long)tg[o] * (mid - l + 1);
sum[o << 1 | 1] += (long)tg[o] * (r - mid);
tg[o] = 0;
}
if(ql <= mid) modify(o << 1, l, mid);
if(mid + 1 <= qr) modify(o << 1 | 1, mid + 1, r);
min[o] = Math.min(min[o << 1], min[o << 1 | 1]);
sum[o] = sum[o << 1] + sum[o << 1 | 1];
}
}
public static void main(String[] args) throws IOException {
String s = in.readLine(); String[] ss = s.split(" ");
int n = Integer.parseInt(ss[0]), k = Integer.parseInt(ss[1]);
s = in.readLine(); ss = s.split(" "); a = new int[n + 1];
for(int i = 1; i <= n; ++i) a[i] = Integer.parseInt(ss[i - 1]);
tg = new int[n << 2]; min = new int[n << 2]; sum = new long[n << 2];
long ans = 0; build(1, 1, n);
for(int i = 1; i + k - 1 <= n; ++i) {
ql = i; qr = i + k - 1;
int res = getMin(1, 1, n);
if(res > 0) {
ans += res; x = -res;
modify(1, 1, n);
}
}
ans += sum[1];
out.write(String.format("%d\n", ans));
out.flush();
}
}