題目鏈接：http://codeforces.com/problemset/problem/815/A

題目：

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

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:

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.

題意：給定n x m的初始零矩陣，每次操作可以為某一行，某一列全部+1。問最小需要步數變成題目中給定的矩陣（如果可以的話），否則輸出-1。

題解：暴力模擬一下過程。先行再列或先列再行。比較一下，最後輸出就可以了。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define PI acos(-1.0)
 5 #define INF 0x3f3f3f3f
 6 #define FAST_IO ios::sync_with_stdio(false)
 7 #define CLR(arr,val) memset(arr,val,sizeof(arr))
 8 const int N=111;
 9 const int D=555;
10 
11 typedef long long LL;
12 int M1[N][N],M2[N][N],r1[N],c1[N],r2[N],c2[N];
13 string ans1[D*D],ans3[D*D];
14 int ans2[D*D],ans4[D*D];
15 
16 int main(){
17     int n,m,cnt1=1,cnt2=1;
18     cin>>n>>m;
19     for(int i=1;i<=n;i++)
20     for(int j=1;j<=m;j++)
21     cin>>M1[i][j],M2[i][j]=M1[i][j];
22 
23     for(int i=1;i<=n;i++){
24         r1[i]=INF;
25         for(int j=1;j<=m;j++) r1[i]=min(r1[i],M1[i][j]);
26         for(int j=1;j<=m;j++) M1[i][j]-=r1[i];
27         if(r1[i]!=0){
28             while(r1[i]--){
29                 ans1[cnt1]="row ";
30                 ans2[cnt1]=i;
31                 cnt1++;
32             }
33         }
34     }
35     for(int i=1;i<=m;i++){
36         int idx=0;
37         c1[i]=INF;
38         for(int j=1;j<=n;j++) c1[i]=min(c1[i],M1[j][i]);
39         for(int j=1;j<=n;j++){
40             M1[j][i]-=c1[i];
41             if(M1[j][i]!=0) idx=1;
42         }
43         if(c1[i]!=0){
44             while(c1[i]--){
45                 ans1[cnt1]="col ";
46                 ans2[cnt1]=i;
47                 cnt1++;
48             }
49         }
50         if(idx==1) {cout<<-1<<endl;return 0;}
51     }
52 
53 
54     for(int i=1;i<=m;i++){
55         c2[i]=INF;
56         for(int j=1;j<=n;j++) c2[i]=min(c2[i],M2[j][i]);
57         for(int j=1;j<=n;j++) M2[j][i]-=c2[i];
58         if(c2[i]!=0){
59             while(c2[i]--){
60                 ans3[cnt2]="col ";
61                 ans4[cnt2]=i;
62                 cnt2++;
63             }
64         }
65     }
66 
67     for(int i=1;i<=n;i++){
68         int idx=0;
69         r2[i]=INF;
70         for(int j=1;j<=m;j++) r2[i]=min(r2[i],M2[i][j]);
71         for(int j=1;j<=m;j++){
72             M2[i][j]-=r2[i];
73             if(M2[i][j]!=0) idx=1;
74         }
75         if(r2[i]!=0){
76             while(r2[i]--){
77                 ans3[cnt2]="row ";
78                 ans4[cnt2]=i;
79                 cnt2++;
80             }
81         }
82         if(idx==1) {cout<<-1<<endl;return 0;}
83     }
84     if(cnt1<=cnt2){
85         cout<<cnt1-1<<endl;
86         for(int i=1;i<cnt1;i++){
87             cout<<ans1[i]<<" "<<ans2[i]<<endl;
88         }
89     }
90     else{
91         cout<<cnt2-1<<endl;
92         for(int i=1;i<cnt2;i++){
93             cout<<ans3[i]<<" "<<ans4[i]<<endl;
94         }
95     }
96     return 0;
97 }

CF 815 A. Karen and Game（暴力，模擬）