Codeforces Round #419 (Div. 2) C. Karen and Game
On the way to school, Karen became fixated on the puzzle game on her phone!
![](http://codeforces.com/predownloaded/13/27/1327478902a073d7774242480818f1d4d93d7270.png)
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 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1
4
row 1
row 1
col 4
row 3
3 3
0 0 0
0 1 0
0 0 0
-1
3 3
1 1 1
1 1 1
1 1 1
3
row 1
row 2
row 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:
![](http://codeforces.com/predownloaded/dd/0f/dd0fcb1e5b1fe4c79e563210a46087f050e31b23.png)
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:
![](http://codeforces.com/predownloaded/e5/66/e56682602939194f6486566eb2bf1a2c106dca6b.png)
Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.
因为是输出任意一组,而且只能+1,所以倒着每行每列考虑,看n和m谁小,若n小先把行处理完在处理列,反之同理。
#include <iostream> #include <cstring> #include <cmath> #include <cstdio> #include <string> #include <algorithm> #include <queue> #include <vector> #include <set> #define inf 99999999 using namespace std; typedef long long ll; const int maxn = 100 + 10; int a[maxn][maxn]; struct node { string s; int x; node(string ss, int xx): s(ss), x(xx) {} }; vector<node> v; int n, m; int main() { while (cin >> n >> m) { v.clear(); bool flag = false; for (int i = 1 ; i <= n ; i ++) { for (int j = 1 ; j <= m ; j ++) { cin >> a[i][j]; if (a[i][j] != 0)flag = true; } } if (flag == false) {cout << "0" << endl; continue;} for (int i = 1 ; i <= n ; i++) { a[i][0] = inf; for (int j = 1 ; j <= m ; j ++) { a[i][0] = min(a[i][0], a[i][j]); } } for (int j = 1 ; j <= m ; j ++) { a[0][j] = inf; for (int i = 1 ; i <= n ; i ++) { a[0][j] = min(a[0][j], a[i][j]); } } if (n <= m) { for (int i = 1 ; i <= n ; i++) { if (a[i][0] == 0)continue; for (int k = a[i][0] ; k >= 1 ; k --) { v.push_back(node("row", i)); } for (int j = 1 ; j <= m ; j ++) { a[0][j] = min(a[0][j], a[i][j] - a[i][0]); } for (int j = 1 ; j <= m ; j ++) { a[i][j] -= a[i][0]; } } for (int j = 1 ; j <= m ; j++) { if (!a[0][j])continue; for (int k = 1 ; k <= a[0][j] ; k++) { v.push_back(node("col", j)); } for (int i = 1 ; i <= n ; i ++) { a[i][j] -= a[0][j]; } } } else { for (int j = 1 ; j <= m ; j++) { if (!a[0][j])continue; for (int k = 1 ; k <= a[0][j] ; k++) { v.push_back(node("col", j)); //cout << "col" << " " << j << endl; } for(int i = 1 ; i <= n ; i ++){ a[i][0] = min(a[i][0],a[i][j]-a[0][j]); } for (int i = 1 ; i <= n ; i ++) { a[i][j] -= a[0][j]; } } for (int i = 1 ; i <= n ; i++) { if (a[i][0] == 0)continue; for (int k = a[i][0] ; k >= 1 ; k --) { v.push_back(node("row", i)); } for (int j = 1 ; j <= m ; j ++) { a[i][j] -= a[i][0]; } } } bool ok = false; for (int i = 1 ; i <= n ; i ++) { for (int j = 1 ; j <= m ; j ++) { if (a[i][j] != 0)ok = true; } } /*for(int i = 1 ; i <= n ; i ++){ for(int j = 1 ; j <= m ; j ++){ cout << a[i][j] << " "; } cout << endl; }*/ if (ok) {cout << "-1" << endl; continue;} if (v.size() == 0 && flag == true) {cout << "-1" << endl; continue;} cout << v.size() << endl; for (int i = v.size() - 1 ; i >= 0 ; i --) { cout << v[i].s << " " << v[i].x << endl; } } return 0; }