codeforces 816C Karen and Game

C. Karen and Game

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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.

找到每行每列的最小值依次消除,但是比较先考虑行和先考虑列哪种更优.

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int n,m;const int MAXN=110;int a[MAXN][MAXN];int aa[MAXN][MAXN];struct node{    int op;    int x;    int cnt;    node(){}    node(int opp,int xx,int yy){        op=opp;        x=xx;        cnt=yy;    }}p[20010],pp[20010];const int INF=100000;int main(){    scanf("%d%d",&n,&m);    int tot=0;    for(int i=1;i<=n;i++){        for(int j=1;j<=m;j++){            scanf("%d",&a[i][j]);            aa[i][j]=a[i][j];        }    }    for(int i=1;i<=n;i++){        int MIN=INF;        for(int j=1;j<=m;j++){            MIN=min(MIN,a[i][j]);        }        if(MIN>0){            p[tot++]=node(0,i,MIN);            for(int j=1;j<=m;j++){                a[i][j]-=MIN;            }        }    }    for(int j=1;j<=m;j++){        int MIN=INF;        for(int i=1;i<=n;i++){            MIN=min(MIN,a[i][j]);        }        if(MIN>0){            p[tot++]=node(1,j,MIN);            for(int i=1;i<=n;i++){                a[i][j]-=MIN;            }        }    }    int to=0;    for(int j=1;j<=m;j++){        int MIN=INF;        for(int i=1;i<=n;i++){            MIN=min(MIN,aa[i][j]);        }        if(MIN>0){            pp[to++]=node(1,j,MIN);            for(int i=1;i<=n;i++){                aa[i][j]-=MIN;            }        }    }    for(int i=1;i<=n;i++){        int MIN=INF;        for(int j=1;j<=m;j++){            MIN=min(MIN,aa[i][j]);        }        if(MIN>0){            pp[to++]=node(0,i,MIN);            for(int j=1;j<=m;j++){                aa[i][j]-=MIN;            }        }    }    int ok=1;    for(int i=1;i<=n;i++){        for(int j=1;j<=m;j++){            if(a[i][j]){                ok=-1;                break;            }        }        if(ok==-1)            break;    }    if(ok==-1){        puts("-1");        return 0;    }    int sum=0;    for(int i=0;i<tot;i++)        sum+=p[i].cnt;    int ssum=0;    for(int i=0;i<to;i++)        ssum+=pp[i].cnt;    if(sum<=ssum)        ok=0;    else        ok=1;    printf("%d\n",min(sum,ssum));    if(ok==0){    for(int i=0;i<tot;i++){        if(p[i].op==0){            for(int j=0;j<p[i].cnt;j++)            printf("row %d\n",p[i].x);        }else{            for(int j=0;j<p[i].cnt;j++)            printf("col %d\n",p[i].x);        }    }    }else{        for(int i=0;i<to;i++){            if(pp[i].op==0){                for(int j=0;j<pp[i].cnt;j++)                    printf("row %d\n",pp[i].x);            }else{                for(int j=0;j<pp[i].cnt;j++)                    printf("col %d\n",pp[i].x);            }        }    }}