【蓝桥杯 最优清零方案 线段树做法】

原题链接

最优清零方案

解题思路

1. 使用线段树进行区间修改和区间查询。
2. 假设全部进行操作1，需要的操作次数为数列和；每使用一次操作2，相较于全部使用操作1减少的操作次数为k-1;

具体代码

import java.io.*;

public class Main {
	public static int N, K, MAX = 1_000_005;
	public static int[] arr = new int[MAX];
	public static long ans;

	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StreamTokenizer in = new StreamTokenizer(br);
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

		ans = 0;
		while (in.nextToken() != StreamTokenizer.TT_EOF) {
			N = (int) in.nval;
			in.nextToken();
			K = (int) in.nval;
			for (int i = 1; i <= N; i++) {
				in.nextToken();
				arr[i] = (int) in.nval;
				ans += arr[i];
			}
		}

		out.println(solve());

		br.close();
		out.flush();
		out.close();
	}

	public static long solve() {
		build(1, N, 1);
		for (int i = 1, j = i + K - 1; j <= N; i++, j++) {
			int tempMin = queryMin(i, j, 1, N, 1);
			if (tempMin > 0) {
				add(i, j, -tempMin, 1, N, 1);
				ans -= tempMin * K - tempMin;
			}
		}
		return ans;
	}

	public static int[] min = new int[MAX << 2];
	public static int[] max = new int[MAX << 2];
	public static int[] add = new int[MAX << 2];

	public static int f(int l, int r, int i) {
		if (max[i] == 0) {
			return 0;
		}
		down(i);
		if (l == r) {
			return max[i];
		}
		int res = 0;
		int mid = (l + r) / 2;
		res += f(l, mid, i << 1) + f(mid + 1, r, i << 1 | 1);
		return res;
	}

	public static void build(int l, int r, int i) {
		if (l > r)
			return;
		add[i] = 0;
		if (l == r) {
			min[i] = arr[l];
			max[i] = arr[l];
		} else {
			int mid = (l + r) / 2;
			build(l, mid, i << 1);
			build(mid + 1, r, i << 1 | 1);
			up(i);
		}
	}

	public static void up(int i) {
		min[i] = Math.min(min[i << 1], min[i << 1 | 1]);
		max[i] = Math.max(max[i << 1], max[i << 1 | 1]);
	}

	public static void down(int i) {
		if (add[i] != 0) {
			lazyAdd(i << 1, add[i]);
			lazyAdd(i << 1 | 1, add[i]);
			add[i] = 0;
		}
	}

	public static void lazyAdd(int i, int v) {
		min[i] += v;
		max[i] += v;
		add[i] += v;
	}

	public static int queryMin(int jobl, int jobr, int l, int r, int i) {
		if (jobl <= l && r <= jobr) {
			return min[i];
		}
		down(i);
		int mid = (l + r) / 2;
		int ans = Integer.MAX_VALUE;
		if (jobl <= mid) {
			ans = Math.min(ans, queryMin(jobl, jobr, l, mid, i << 1));
		}
		if (mid + 1 <= jobr) {
			ans = Math.min(ans, queryMin(jobl, jobr, mid + 1, r, i << 1 | 1));
		}
		return ans;
	}

	public static void add(int jobl, int jobr, int jobv, int l, int r, int i) {
		if (jobl <= l && r <= jobr) {
			lazyAdd(i, jobv);
		} else {
			down(i);
			int mid = (l + r) / 2;
			if (jobl <= mid) {
				add(jobl, jobr, jobv, l, mid, i << 1);
			}
			if (jobr > mid) {
				add(jobl, jobr, jobv, mid + 1, r, i << 1 | 1);
			}
			up(i);
		}
	}
}