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!
Input
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).
Output
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.
Examples
input
3 52 2 2 3 20 0 0 1 01 1 1 2 1
output
4row 1row 1col 4row 3
input
3 30 0 00 1 00 0 0
output
-1
input
3 31 1 11 1 11 1 1
output
3row 1row 2row 3
Note
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 }
原文地址：https://www.cnblogs.com/Leonard-/p/8110938.html