Codeforces 816C Karen and Game【思维】

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 number0.

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 tog_i, 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 andm (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 thei-th of these rows contains g_i, j (0 ≤ g_i, 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 thex-th row".
• col x, (1 ≤ x ≤ m) describing a move of the form "choose thex-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 and5 columns. She can perform the following 4 moves to beat the level:

In the second test case, Karen has a grid with 3 rows and3 columns. It is clear that it is impossible to beat the level; performing any move will create three1s on the grid, but it is required to only have one1 in the center.

In the third test case, Karen has a grid with 3 rows and3 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.

题目大意：

每次操作可以选择一行或者一列进行整体+1的操作，问最少步数，使得一开始的全0矩阵变成输入进来的N*M的矩阵。

思路：

直接模拟，先行后列的去做，每行取最小值，以及每列取最小值，那么就是这一行和这一列的操作数量。

对应模拟出来就行了。

因为是要最小步数，所以有可能要先列后行的去做，那么我们两种情况都做以下然后比较操作数量即可。

Ac代码（代码稍微有点挫）：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int c[150][150];int d[150][150];int ans[100000][3];int ans2[100000][3];int a[150][150];int b[150][150];int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        int output=0;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                scanf("%d",&a[i][j]);                c[i][j]=a[i][j];            }        }        int cnt=0;        for(int i=1;i<=n;i++)        {            int minn=0x3f3f3f3f;            for(int j=1;j<=m;j++)            {                minn=min(minn,a[i][j]);            }            if(minn==0)continue;            output+=minn;            ans[cnt][0]=0;            ans[cnt][1]=i;            ans[cnt++][2]=minn;            for(int j=1;j<=m;j++)            {                b[i][j]+=minn;                a[i][j]-=minn;            }        }        for(int j=1;j<=m;j++)        {            int minn=0x3f3f3f3f;            for(int i=1;i<=n;i++)            {                minn=min(minn,a[i][j]);            }            if(minn==0)continue;            output+=minn;            ans[cnt][0]=1;            ans[cnt][1]=j;            ans[cnt++][2]=minn;            for(int i=1;i<=n;i++)            {                b[i][j]+=minn;                a[i][j]-=minn;            }        }        int flag=0;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                if(a[i][j]!=0)flag=1;            }        }        if(flag==0)        {            int output2=0;            for(int i=1;i<=n;i++)            {                for(int j=1;j<=m;j++)                {                    a[i][j]=c[i][j];                    b[i][j]=0;                }            }            int cnt2=0;            for(int j=1;j<=m;j++)            {                int minn=0x3f3f3f3f;                for(int i=1;i<=n;i++)                {                    minn=min(minn,a[i][j]);                }                if(minn==0)continue;                output2+=minn;                ans2[cnt2][0]=1;                ans2[cnt2][1]=j;                ans2[cnt2++][2]=minn;                for(int i=1;i<=n;i++)                {                    b[i][j]+=minn;                    a[i][j]-=minn;                }            }            for(int i=1;i<=n;i++)            {                int minn=0x3f3f3f3f;                for(int j=1;j<=m;j++)                {                    minn=min(minn,a[i][j]);                }                if(minn==0)continue;                output2+=minn;                ans2[cnt2][0]=0;                ans2[cnt2][1]=i;                ans2[cnt2++][2]=minn;                for(int j=1;j<=m;j++)                {                    b[i][j]+=minn;                    a[i][j]-=minn;                }            }            if(output<=output2)            {                printf("%d\n",output);                for(int i=0;i<cnt;i++)                {                    for(int j=0;j<ans[i][2];j++)                    {                        if(ans[i][0]==0)printf("row ");                        else printf("col ");                        printf("%d\n",ans[i][1]);                    }                }            }            else            {                printf("%d\n",output2);                for(int i=0;i<cnt2;i++)                {                    for(int j=0;j<ans2[i][2];j++)                    {                        if(ans2[i][0]==0)printf("row ");                        else printf("col ");                        printf("%d\n",ans2[i][1]);                    }                }            }        }        else printf("-1\n");    }}