C. 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
枚举暴力,一开始没考虑n>m的情况,应该是以列为主的。直接GG掉了QAQ QAQ QAQ QAQ还是功夫不到家啊。
#include<bits/stdc++.h>using namespace std;const int MAXN=100+7;const int inf = 1e9;int n,m;int g[MAXN][MAXN];int r[MAXN],c[MAXN];int main(){ scanf("%d%d",&n,&m); for(int i = 1 ; i <= n ;++i) for(int j = 1 ; j <= m ;++j)scanf("%d",&g[i][j]); int flag; if(n<m) { for(int i = 1 ; i <= n ; ++i) { r[i] = inf; for(int j = 1; j <= m ; ++j) { r[i] = min(r[i],g[i][j]); } } flag = 1; for(int i = 1 ; i <= n ; ++i)if(flag) { for(int j = 1 ; j <= m ; ++j) { if(g[i][j] >= r[i]+c[j]) { c[j] += g[i][j] - r[i] - c[j]; } else { flag = 0; break; } } } for(int i = 1; i <= n ; ++i)if(flag) for(int j = 1 ; j <= m ; ++j) { if(g[i][j] != r[i]+c[j]) { flag = 0; break; } } } else { for(int j = 1 ; j <= m ; ++j) { c[j] = inf; for(int i = 1; i <= n ; ++i) { c[j] = min(c[j],g[i][j]); } } flag = 1; for(int j = 1 ; j <= m ; ++j)if(flag) { for(int i = 1 ; i <= n ; ++i) { if(g[i][j] >= r[i]+c[j]) { r[i] += g[i][j] - r[i] - c[j]; } else { flag = 0; break; } } } for(int i = 1; i <= n ; ++i)if(flag) for(int j = 1 ; j <= m ; ++j) { if(g[i][j] != r[i]+c[j]) { flag = 0; break; } } } if(!flag) { puts("-1"); return 0; } int ans = 0; for(int i = 1 ; i <= n ; ++i)ans += r[i]; for(int i = 1 ; i <= m ; ++i)ans += c[i]; printf("%d\n",ans); for(int i = 1 ; i <= n ; ++i) for(int j = 0 ; j < r[i] ; ++j)printf("row %d\n",i); for(int i = 1 ; i <= m ; ++i) for(int j = 0 ; j < c[i] ; ++j)printf("col %d\n",i);}
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