codeforces815A Karen and Game
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On the way to school, Karen became fixated on the puzzle game on her phone!
The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.
One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.
To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.
Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!
The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.
The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).
If there is an error and it is actually not possible to beat the level, output a single integer -1.
Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.
The next k lines should each contain one of the following, describing the moves in the order they must be done:
- row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row".
- col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column".
If there are multiple optimal solutions, output any one of them.
3 52 2 2 3 20 0 0 1 01 1 1 2 1
4row 1row 1col 4row 3
3 30 0 00 1 00 0 0
-1
3 31 1 11 1 11 1 1
3row 1row 2row 3
In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:
In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.
In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:
Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.
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题目的意思是给出一个全0矩阵,操作是在一行或一列全+1,问怎样才能得到给定矩阵
思路:从给定矩阵出发,每一行每一列能全减就贪心减去,根据行列数量决定先处理行还是列
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <cmath>using namespace std;#define LL long longconst int inf=0x7fffffff;int mp[500][500];struct node{ int f,x,nu;} ans[200000];int n,m,cnt,tot;void Row(){ for(int i=0; i<m; i++) { int mn=inf; for(int j=0; j<n; j++) { mn=min(mn,mp[i][j]); } for(int j=0; j<n; j++) { mp[i][j]-=mn; } if(mn>0) { tot+=mn; ans[cnt].f=1; ans[cnt].x=mn; ans[cnt++].nu=i+1; } }}void Col(){ for(int j=0; j<n; j++) { int mn=inf; for(int i=0; i<m; i++) { mn=min(mn,mp[i][j]); } for(int i=0; i<m; i++) { mp[i][j]-=mn; } if(mn>0) { tot+=mn; ans[cnt].f=2; ans[cnt].x=mn; ans[cnt++].nu=j+1; } }}int main(){ scanf("%d%d",&m,&n); for(int i=0; i<m; i++) for(int j=0; j<n; j++) scanf("%d",&mp[i][j]); cnt=0; tot=0; if(m>n) { Col(); Row(); } else { Row(); Col(); }int flag=0; for(int i=0; i<m; i++) for(int j=0; j<n; j++) { if(mp[i][j]>0) { flag=1; break; } if(flag) break; } if(flag) printf("-1\n"); else { printf("%d\n",tot); for(int i=0; i<cnt; i++) { if(ans[i].f==1) for(int j=0;j<ans[i].x;j++) printf("row %d\n",ans[i].nu); else for(int j=0;j<ans[i].x;j++) printf("col %d\n",ans[i].nu); } } return 0;}
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