[Cometoj#4 C]方块切割_质因数分解_贪心

方块切割

题目链接：https://cometoj.com/contest/39/problem/C?problem_id=1583

数据范围：略。

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题解：

首先，如果我们知道了多少道在行上，多少刀在列上，应该怎么办？

不难发现行和列是独立的，只需要分别保证各自均分了网格即可。

那么怎么切呢？只需要顺次枚举，能切就切即可。

至于怎么知道行和列各自切多少刀？

枚举呗

代码：

#include <bits/stdc++.h>

#define N 1010 

using namespace std;

int r[N], c[N];
char a[N][N];
int s[N][N];
int ans[2 * N];
int x[N], y[N];
int n, m, k;

int main() {
	int T;
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d%d", &n, &m, &k);
		for (int i = 1; i <= n; i++) {
			r[i] = 0;
		}
		for (int i = 1; i <= m; i++) {
			c[i] = 0;
		}
		int sum = 0;
		for (int i = 1; i <= n; i++) {
			scanf("%s", a[i] + 1);
			for (int j = 1; j <= m; j++) {
				if (a[i][j] == '0') {
					r[i]++;
					c[j]++;
					sum++;
				}
			}
		}
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];
				s[i][j] += a[i][j] == '0';
			}
		}
		for (int i = 1; i <= k; i++) {
			ans[i] = n + m - 1;
		}
		if (sum == 0) {
			for (int i = 1; i < k; i++) {
				printf("%d ", i);
			}
			printf("%d", k);
			printf("\n");
			continue;
		}
		for (int l = k; l >= 0; l--) {
			if (sum % ((l + 1) * (k - l + 1))) continue;
			int br = sum / (l + 1);
			int bc = sum / (k - l + 1);
			int b = sum / (l + 1) / (k - l + 1);
			bool f = 1;
			int cur = 0;
			int X = 0;
			for (int i = 1; i <= n; i++) {
				cur += r[i];
				if (cur == br) {
					x[++X] = i;
					cur = 0;
				}
				if (cur > br) f = 0;
			}
			cur = 0;
			int Y = 0;
			for (int i = 1; i <= m; i++) {
				cur += c[i];
				if (cur == bc) {
					y[++Y] = i;
					cur = 0;
				}
				if (cur > bc) f = 0;
			}
			for (int i = 1; i <= X; i++) {
				for (int j = 1; j <= Y; j++) {
					if (s[x[i]][y[j]] - s[x[i - 1]][y[j]] - s[x[i]][y[j - 1]] + s[x[i - 1]][y[j - 1]] != b) f = 0;
				}
			}
			if (!f) continue;
			for (int i = 1; i < X; i++) ans[i] = x[i];
			for (int i = 1; i < Y; i++) ans[l + i] = y[i] + n - 1;
			break;
		}
		if (ans[1] == n + m - 1) {
			printf("Impossible\n");
		}
		else {
			for (int i = 1; i < k; i++) {
				printf("%d ", ans[i]);
			}
			printf("%d", ans[k]);
			printf("\n");
		}
	}
	return 0;
}