Round#419 C Karen and Game

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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!

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 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1
output
4
row 1
row 1
col 4
row 3

input
3 3
0 0 0
0 1 0
0 0 0
output
-1

input
3 3
1 1 1
1 1 1
1 1 1
output
3
row 1
row 2
row 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.
一道模拟题，当时看题没反应过来，居然project accept 了…..测试数据是专门用来卡人的么…..
然后大力xjb瞎搞…..
附上AC代码：

#include<bits/stdc++.h>using namespace std;const int INF=0x3f3f3f3f;int main(){    ios::sync_with_stdio(0);    cin.tie(0);    int n,m;    cin>>n>>m;    int ans[105][105]={0},ar[105][105]={0};    for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++)        {cin>>ans[i][j];ar[i][j]=ans[i][j];}    int mins=INF,sum=0;    int temp[6][200]={0};    for(int i=1;i<=n;i++) {        mins=INF;        for(int j=1;j<=m;j++)            mins=min(mins,ans[i][j]);        temp[1][i]=mins;        for(int j=1;j<=m;j++)            ans[i][j]-=mins;    }    for(int j=1;j<=m;j++){        mins=INF;        for(int i=1;i<=n;i++)            mins=min(mins,ans[i][j]);        temp[2][j]=mins;        for(int i=1;i<=n;i++)            ans[i][j]-=mins;    }    for(int j=1;j<=m;j++){        mins=INF;        for(int i=1;i<=n;i++)            mins=min(mins,ar[i][j]);        temp[3][j]=mins;        for(int i=1;i<=n;i++)            ar[i][j]-=mins;    }    for(int i=1;i<=n;i++) {        mins=INF;        for(int j=1;j<=m;j++)            mins=min(mins,ar[i][j]);        temp[4][i]=mins;        for(int j=1;j<=m;j++)            ar[i][j]-=mins;    }    int flag1=0,flag2=0;    for(int i=1;i<=n;i++) {        for(int j=1;j<=m;j++)  {            if(ans[i][j]!=0) {            flag1=1;            break;            }        }    }    for(int i=1;i<=n;i++) {        for(int j=1;j<=m;j++)  {            if(ar[i][j]!=0) {            flag2=1;            break;            }        }    }    if(flag1==1&&flag2==1) cout<<-1<<endl;    else if(flag1==0&&flag2==0) {        int sum1=0,sum2=0;        for(int i=1;i<=n;i++) {            sum1+=temp[1][i];            sum2+=temp[4][i];        }        for(int j=1;j<=m;j++) {            sum1+=temp[2][j];            sum2+=temp[3][j];        }        if(sum1>sum2) {            cout<<sum2<<endl;            for(int i=1;i<=n;i++)                for(int j=1;j<=temp[4][i];j++)                    cout<<"row "<<i<<endl;            for(int j=1;j<=m;j++)                for(int i=1;i<=temp[3][j];i++)                     cout<<"col "<<j<<endl;        }        else {            cout<<sum1<<endl;            for(int i=1;i<=n;i++)                for(int j=1;j<=temp[1][i];j++)                    cout<<"row "<<i<<endl;            for(int j=1;j<=m;j++)                for(int i=1;i<=temp[2][j];i++)                     cout<<"col "<<j<<endl;        }    }    else if(flag1==1) {        int sum2=0;        for(int i=1;i<=n;i++) {            sum2+=temp[4][i];        }        for(int j=1;j<=m;j++) {            sum2+=temp[3][j];        }        cout<<sum2<<endl;            for(int i=1;i<=n;i++)                for(int j=1;j<=temp[4][i];j++)                    cout<<"row "<<i<<endl;            for(int j=1;j<=m;j++)                for(int i=1;i<=temp[3][j];i++)                     cout<<"col "<<j<<endl;    }    else if(flag2==1) {        int sum1=0;        for(int i=1;i<=n;i++) {            sum1+=temp[4][i];        }        for(int j=1;j<=m;j++) {            sum1+=temp[3][j];        }        cout<<sum1<<endl;            for(int i=1;i<=n;i++)                for(int j=1;j<=temp[1][i];j++)                    cout<<"row "<<i<<endl;            for(int j=1;j<=m;j++)                for(int i=1;i<=temp[2][j];i++)                     cout<<"col "<<j<<endl;    }    return 0;}