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USACO 2021 February Contest, Silver Problem 3. Just Green Enough

Category: 国际竞赛, 计算机国际竞赛 Date: 2022年7月4日下午5:44

USACO 2021 February Contest, Silver Problem 3. Just Green Enough

Farmer John's pasture can be regarded as an $N \times N$ grid ( $1 \leq N \leq 500$ ) of square "cells" of grass (picture a huge chessboard). Due to soil variability, the grass in some cells is greener than in others. Each cell $(i, j)$ is described by an integer level of green-ness $G (i, j)$ , ranging from $1 \dots 200$ .

Farmer John wants to take a photograph of a rectangular sub-grid of his pasture. He wants to be sure the sub-grid looks sufficiently green, but not ridiculously green, so he decides to photograph a sub-grid for which the minimum value of $G$ is exactly 100. Please help him determine how many different photographs he could possibly take. A sub-grid can be as large as the entire pasture or as small as a single grid cell (there are $N^{2} (N + 1)^{2} / 4$ different sub-grids in total --- note that this number might be too large to store in a standard 32-bit integer, so you might need to use 64-bit integer data types like a "long long" in C++).

INPUT FORMAT (input arrives from the terminal / stdin):

The first line of input contains $N$ . The next $N$ lines each contain $N$ integers and collectively describe the $G (i, j)$ values for the $N \times N$ pasture.

OUTPUT FORMAT (print output to the terminal / stdout):

Please print the number of distinct photos Farmer John can take -- that is, the number of rectangular sub-grids for which the minimum level of green-ness is exactly 100.Note that the large size of integers involved in this problem may require the use of 64-bit integer data types (e.g., a "long long" in C/C++).

SAMPLE INPUT:

3
57 120 87
200 100 150
2 141 135

SAMPLE OUTPUT:

SCORING:

Test cases 1-5 satisfy $N \leq 200$ .
Test cases 6-10 satisfy no additional constraints.

Problem credits: Brian Dean

USACO 2021 February Contest, Silver Problem 3. Just Green Enough 题解(翰林国际教育提供，仅供参考)

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(Analysis by Benjamin Qi)

The first step is to use complementary counting. The number of rectangular sub-grids with minimum equal to $100$ is equal to the number of rectangular sub-grids with minimum at least $100$ minus the number of rectangular sub-grids with minimum at least $101$ .

To count the number of rectangular sub-grids with minimum at least $m$ , create an $N \times N$ boolean array $o k$ such that $o k [i] [j] = 1$ if $G [i] [j] \geq m$ . We want to count the number of rectangular sub-grids in $o k$ that consist solely of ones.

If $o k$ was an $N \times 1$ rectangle rather than an $N \times N$ rectangle, the following loop would suffice to compute the answer:

int run = 0;
for (int i = 0; i < N; ++i) {
	if (ok[i][0]) ans += ++run;
	else run = 0;
}

Each run of $l$ consecutive ones contributes $\frac{l (l + 1)}{2}$ to the answer.

Define ${all_ones}_{i, j} [k]$ to be true if all of the cells from $(i, k)$ to $(j, k)$ contain ones, and false otherwise. It suffices to iterate over $(i, j)$ , compute ${all_ones}_{i, j} [k]$ for all $0 \leq k < N$ , and then apply the 1D solution to $all_ones$ . This takes $O (N^{4})$ time since there are $O (N^{3})$ triples $(i, j, k)$ and for each one, we do $O (N)$ work to compute ${all_ones}_{i, j} [k]$ .

To speed this up to $O (N^{3})$ time, we can use 1D prefix sums to compute ${all_ones}_{i, j} [k]$ in $O (1)$ time.

Danny Mittal's code:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
 
public class JustGreenEnough2 {
 
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(in.readLine());
        int[][] pasture = new int[n][n];
        for (int y = 0; y < n; y++) {
            StringTokenizer tokenizer = new StringTokenizer(in.readLine());
            for (int x = 0; x < n; x++) {
                pasture[y][x] = Integer.parseInt(tokenizer.nextToken());
            }
        }
        int[][] sumsBelow = new int[n][n + 1];
        int[][] sumsAtMost = new int[n][n + 1];
        for (int y = 0; y < n; y++) {
            for (int x = 0; x < n; x++) {
                sumsBelow[y][x + 1] = sumsBelow[y][x] + (pasture[y][x] < 100 ? 1 : 0);
                sumsAtMost[y][x + 1] = sumsAtMost[y][x] + (pasture[y][x] <= 100 ? 1 : 0);
            }
        }
        long answer = 0;
        for (int x1 = 0; x1 < n; x1++) {
            for (int x2 = x1 + 1; x2 <= n; x2++) {
                int y1 = -1;
                int y2 = -1;
                for (int y0 = 0; y0 < n; y0++) {
                    while (y1 < n && (y1 < y0 || sumsAtMost[y1][x2] - sumsAtMost[y1][x1] == 0)) {
                        y1++;
                    }
                    while (y2 < n && (y2 < y0 || sumsBelow[y2][x2] - sumsBelow[y2][x1] == 0)) {
                        y2++;
                    }
                    answer += y2 - y1;
                }
            }
        }
        System.out.println(answer);
    }
}

Alternatively, note that ${all_ones}_{i, j} [k] = {all_ones}_{i, j - 1} [k] & o k [j] [k]$ . So let's fix $i$ and compute

all_ones i, i, all_ones i, i + 1, \dots, all_ones i, N - 1

in order.

#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
 
int N;
bool ok[1000][1000];
 
ll solve() {
	ll ans = 0;
	for (int i = 0; i < N; ++i) {
		vector<bool> all_ones(N,true);
		for (int j = i; j < N; ++j) { 
			// add rectangles with upper row i and lower row j
			int run = 0;
			for (int k = 0; k < N; ++k) {
				// all_ones_{i,j-1}[k] -> all_ones_{i,j}[k]
				all_ones[k] = all_ones[k]&ok[j][k]; 
				if (all_ones[k]) ans += ++run; // update answer
				else run = 0;
			}
		}
	}
	return ans;
}
 
int main() {
	cin >> N;
	vector<vector<int>> pasture(N,vector<int>(N)); 
	for (vector<int>& a: pasture) 
		for (int& b: a) cin >> b;
 
	for (int i = 0; i < N; ++i) 
		for (int j = 0; j < N; ++j)
			ok[i][j] = pasture[i][j] >= 100;
	ll ans = solve();
 
	for (int i = 0; i < N; ++i) 
		for (int j = 0; j < N; ++j)
			ok[i][j] = pasture[i][j] > 100;
	ans -= solve();
 
	cout << ans << "\n";
}

It was possible (but not necessary) to solve this problem in $O (N^{2})$ time. In the code below, for a fixed $i$ , I iterate over all $j$ in decreasing (rather than increasing order as the solution above does) and maintain the sum of the contributions of all runs in ${all_ones}_{i, j}$ in $sum_comb$ . When $j$ decreases by one, I update $sum_comb$ accordingly for each $k$ such that ${all_ones}_{i, j + 1} [k] = 0$ and ${all_ones}_{i, j} [k] = 1$ .

#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
 
int N;
bool ok[1000][1000];
 
ll solve() {
	ll ans = 0;
	vector<int> lst(N,N-1);
	vector<int> to_add[1000];
	for (int i = N-1; i >= 0; --i) {
		for (int j = i; j < N; ++j) to_add[j].clear();
		for (int k = 0; k < N; ++k) {
			if (ok[i][k] == 0) lst[k] = i-1;
			else to_add[lst[k]].push_back(k);
		}
		int sum_comb = 0;
		vector<int> lef(N,-1), rig(N,-1);
		for (int j = N-1; j >= i; --j) {
			for (int k: to_add[j]) {
				// all_ones_{i,j+1}[k] = 0, all_ones_{i,j}[k] = 1
				int l = k, r = k;
				auto c2 = [](int x) { return (x+1)*(x+2)/2; };
				if (k && lef[k-1] != -1) {
					l = lef[k-1];
					sum_comb -= c2(k-1-l);
				}
				if (k+1 < N && rig[k+1] != -1) {
					r = rig[k+1];
					sum_comb -= c2(r-k-1);
				}
				lef[r] = l, rig[l] = r;
				sum_comb += c2(r-l);
			}
			ans += sum_comb;
		}
	}
	return ans;
}
 
int main() {
	cin >> N;
	vector<vector<int>> pasture(N,vector<int>(N)); 
	for (vector<int>& a: pasture) 
		for (int& b: a) cin >> b;
 
	for (int i = 0; i < N; ++i) 
		for (int j = 0; j < N; ++j)
			ok[i][j] = pasture[i][j] >= 100;
	ll ans = solve();
 
	for (int i = 0; i < N; ++i) 
		for (int j = 0; j < N; ++j)
			ok[i][j] = pasture[i][j] > 100;
	ans -= solve();
 
	cout << ans << "\n";
}

For an additional challenge, try Maximum Building II.

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