C. Karen and Game

時間 2019-11-16

標籤 karen game 欄目 C&C++ 简体版

原文原文鏈接

On the way to school, Karen became fixated on the puzzle game on her phone!ios

$\text{[math]}$

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.網絡

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.測試

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 g_i, j.ui

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!this

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.spa

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains g_i, j (0 ≤ g_i, j ≤ 500).設計

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.3d

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.code

The next k lines should each contain one of the following, describing the moves in the order they must be done:orm

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

Output

4
row 1
row 1
col 4
row 3

Input

Output

-1

Input

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:

$\text{[math]}$

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:

$\text{[math]}$

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

題解：

這道題首先想過網絡流，然而沒有什麼用。

後來發現貪心能夠過，求出每行每列的最小值，而後一個一個找，從行開始，每次維護一下列的最小值，最後統計一下值的總合是否是和以前的相同，不是的話就說明還有殘餘，如樣例2。是的話就輸出結果。

注：這個代碼有問題，正解在最下面

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
int s[101][101],n,m,sum,sum2;
int mminn[101],mminm[101];
int ans1[101],cnt,ans2[101];
int main()
{
    int i,j;
    memset(mminn,127/3,sizeof(mminn));
    memset(mminm,127/3,sizeof(mminm));
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
        {
            scanf("%d",&s[i][j]);
            sum+=s[i][j];
        }
    }
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
        mminn[i]=min(mminn[i],s[i][j]);
    }
    for(j=1;j<=m;j++)
    {
        for(i=1;i<=n;i++)
        mminm[j]=min(mminm[j],s[i][j]);
    }
    for(i=1;i<=n;i++)
    {
        ans1[i]=mminn[i];
        for(j=1;j<=m;j++)
        {
            s[i][j]-=mminn[i];
            mminm[j]=min(mminm[j],s[i][j]);
        }
    }
    for(j=1;j<=m;j++)
    {
        ans2[j]=mminm[j];
    }
    for(i=1;i<=n;i++)
    {
        sum2+=ans1[i]*m;
    }
    for(i=1;i<=m;i++)
    {
        sum2+=ans2[i]*n;
    }
    if(sum==sum2)
    {
        for(i=1;i<=n;i++)
        if(ans1[i])cnt+=ans1[i];
        for(i=1;i<=m;i++)
        if(ans2[i])cnt+=ans2[i];
        printf("%d\n",cnt);
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=ans1[i];j++)
            printf("row %d\n",i);
        }
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=ans2[i];j++)
            printf("col %d\n",i);
        }
    }
    else cout<<"-1";
    return 0;
}

沒錯，這份代碼確實有問題，很顯然隨便設計一組測試數據

input

4 3

1 1 1

output

col 1

col 2

col 3

而上面這份代碼會輸出

output

row 1

row 2

row 3

row 4

而後就WA掉了,（心痛）。可是修改也很簡單，最後判斷一下是從行仍是從列開始就能夠了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
int s[101][101],n,m,sum,sum2;
int mminn[101],mminm[101];
int ans1[101],cnt,ans2[101];
int main()
{
    int i,j;
    memset(mminn,127/3,sizeof(mminn));
    memset(mminm,127/3,sizeof(mminm));
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
        {
            scanf("%d",&s[i][j]);
            sum+=s[i][j];
        }
    }
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
        mminn[i]=min(mminn[i],s[i][j]);
    }
    for(j=1;j<=m;j++)
    {
        for(i=1;i<=n;i++)
        mminm[j]=min(mminm[j],s[i][j]);
    }
    if(m>n)
    {
        for(i=1;i<=n;i++)
        {
            ans1[i]=mminn[i];
            for(j=1;j<=m;j++)
            {
                s[i][j]-=mminn[i];
                mminm[j]=min(mminm[j],s[i][j]);
            }
        }
        for(j=1;j<=m;j++)
        {
            ans2[j]=mminm[j];
        }
        for(i=1;i<=n;i++)
        {
            sum2+=ans1[i]*m;
            }
        for(i=1;i<=m;i++)
        {
            sum2+=ans2[i]*n;
        }
        if(sum==sum2)
        {
            for(i=1;i<=n;i++)
            if(ans1[i])cnt+=ans1[i];
            for(i=1;i<=m;i++)
            if(ans2[i])cnt+=ans2[i];
            printf("%d\n",cnt);
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=ans1[i];j++)
                printf("row %d\n",i);
            }
            for(i=1;i<=m;i++)
            {
                for(j=1;j<=ans2[i];j++)
                printf("col %d\n",i);
            }
        }
        else cout<<"-1";
    }
    else
    {
        for(i=1;i<=m;i++)
        {
            ans2[i]=mminm[i];
            for(j=1;j<=n;j++)
            {
                s[j][i]-=mminm[i];
                mminn[j]=min(mminn[j],s[j][i]);
            }
        }
        for(j=1;j<=n;j++)
        {
            ans1[j]=mminn[j];
        }
        for(i=1;i<=n;i++)
        {
            sum2+=ans1[i]*m;
        }
        for(i=1;i<=m;i++)
        {
            sum2+=ans2[i]*n;
        }
        if(sum==sum2)
        {
            for(i=1;i<=n;i++)
            if(ans1[i])cnt+=ans1[i];
            for(i=1;i<=m;i++)
            if(ans2[i])cnt+=ans2[i];
            printf("%d\n",cnt);
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=ans1[i];j++)
                printf("row %d\n",i);
            }
            for(i=1;i<=m;i++)
            {
                for(j=1;j<=ans2[i];j++)
                printf("col %d\n",i);
            }
        }
        else cout<<"-1";
    }
    return 0;
}

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