Luogu3444：[POI2006]ORK-Ploughing

题意

见luogu

Sol

贪心+枚举

如果知道最后一次是消除一行

那么一定消了\(n\)行

此时只要消的列最小就好了

枚举每列从上往下消到哪里，那么下面消的越小越好

那么就有了贪心策略：

先消左右的列，再消上面，再消下面

最后一次消列的情况是一样的

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2005);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int k, m, n, s1[_][_], s2[_][_], ans = 2e9;

IL int Calc1(RG int x){
	RG int l1 = 1, r1 = n, l2 = 1, r2 = m, ret = 0;
	while(l1 <= r1 && l2 <= r2){
		++ret;
		if(s2[r1][l2] - s2[l1 - 1][l2] <= k) ++l2;
		else if(s2[r1][r2] - s2[l1 - 1][r2] <= k) --r2;
		else if(s1[l1][r2] - s1[l1][l2 - 1] <= k && l1 < x) ++l1;
		else if(s1[r1][r2] - s1[r1][l2 - 1] <= k) --r1;
		else return 2e9;
	}
	return ret;
}

IL int Calc2(RG int x){
	RG int l1 = 1, r1 = n, l2 = 1, r2 = m, ret = 0;
	while(l1 <= r1 && l2 <= r2){
		++ret;
		if(s1[l1][r2] - s1[l1][l2 - 1] <= k) ++l1;
		else if(s1[r1][r2] - s1[r1][l2 - 1] <= k) --r1;
		else if(s2[r1][l2] - s2[l1 - 1][l2] <= k && l2 < x) ++l2;
		else if(s2[r1][r2] - s2[l1 - 1][r2] <= k) --r2;
		else return 2e9;
	}
	return ret;
}

int main(RG int argc, RG char* argv[]){
	k = Input(), m = Input(), n = Input();
	for(RG int i = 1; i <= n; ++i)
		for(RG int j = 1; j <= m; ++j)
			s1[i][j] = s2[i][j] = Input();
	for(RG int i = 1; i <= n; ++i)
		for(RG int j = 2; j <= m; ++j)
			s1[i][j] += s1[i][j - 1];
	for(RG int j = 1; j <= m; ++j)
		for(RG int i = 2; i <= n; ++i)
			s2[i][j] += s2[i - 1][j];
	for(RG int i = 1; i <= n + 1; ++i) ans = min(ans, Calc1(i));
	for(RG int i = 1; i <= m + 1; ++i) ans = min(ans, Calc2(i));
	printf("%d\n", ans);
	return 0;
}