题目链接：http://codeforces.com/problemset/problem/815/A
题目：
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 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:
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.
题意：给定n x m的初始零矩阵，每次操作可以为某一行，某一列全部+1。问最小需要步数变成题目中给定的矩阵（如果可以的话），否则输出-1。
题解：暴力模拟一下过程。先行再列或先列再行。比较一下，最后输出就可以了。
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 }
