传送门

 

f[i][j] 表示前 i 盆花,放到前 j 个花盆中的最优解

pre[i][j] 记录前驱

#include <cstdio>
#include <cstring>
#include <iostream>
#define N 1001
#define max(x, y) ((x) > (y) ? (x) : (y))

int n, m, t;
int a[N][N], f[N][N], pre[N][N], s[N];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

int main()
{
	int i, j, k;
	n = read();
	m = read();
	for(i = 1; i <= n; i++)
		for(j = 1; j <= m; j++)
			a[i][j] = read();
	for(i = 1; i <= n; i++)
		for(j = i; j <= m; j++)
		{
			f[i][j] = -1e9;
			for(k = i; k <= j; k++)
				if(f[i][j] < f[i - 1][k - 1] + a[i][k])
				{
					f[i][j] = f[i - 1][k - 1] + a[i][k];
					pre[i][j] = k - 1;
				}
		}
	k = m;
	for(i = n; i; i--)
	{
		s[++t] = pre[i][k] + 1;
		k = pre[i][k];
	}
	printf("%d\n", f[n][m]);
	for(i = t; i; i--) printf("%d ", s[i]);
	return 0;
}

  

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