Based on initial observation, it seems that greedily pick the smallest row / column length works. But the last example test case outputs 35 while greedy gives 36.
How you should go from there:
1. check if your greedy implementation is correct. It was, in this case.
2. if no implementation bugs, then maybe greedy does not work, think dp.
3. the input constraint implies a dp of O(N * K * K) is fast enough.
dp[i][j]: min cost to get score j from [1, i] rectangles;
dp[i][j] = min {dp[i - 1][j - currTake] + cost[i][currTaken]}; loop on i and j, for a given score j the current rectangle i, we can take [0, j] scores from rectangle i.
The remaining work is to precompute cost[i][x], the min cost of getting score x from rectangle i.
For a given score x, it is a combination of getting score from completing a row or column.
so we iterate over a rectangle's width and height to compute cost[i][x].
Don't forget to deduct the intersection between completed rows and columns as we do not need to paint these cells twice.
int[][] cost = new int[n + 1][k + 1]; for(int i = 0; i < cost.length; i++) { Arrays.fill(cost[i], Integer.MAX_VALUE); cost[i][0] = 0; } for(int i = 1; i <= n; i++) { for(int r = 0; r <= x[i][0]; r++) { for(int c = 0; c <= x[i][1]; c++) { if(r + c <= k) { cost[i][r + c] = min(cost[i][r + c], r * x[i][1] + c * x[i][0] - r * c); } } } }
标签:Rows,cost,Color,CodeForces,greedy,int,score,rectangle,dp From: https://www.cnblogs.com/lz87/p/18359992