Karen and Game（CodeForces 816C）_codeforces - 816c karen and game(简单模拟)-优快云博客

本文介绍了一种解决特定游戏级别最优解的策略，通过选择行或列并增加数值，目标是在最少的操作次数下达到预设的目标矩阵。文章提供了一个算法实现，首先找到每行和每列的最小值进行调整，然后检查是否所有元素都已达到目标值。

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Description

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:

rowx, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row".
colx, (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.

Sample Input

Input

Output

4
row 1
row 1
col 4
row 3

Input

Output

-1

Input

Output

3
row 1
row 2
row 3

Hint

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.

题解：最优解！！！

代码如下：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#define max(a,b)   (a>b?a:b)
#define min(a,b)   (a<b?a:b)
#define swap(a,b)  (a=a+b,b=a-b,a=a-b)
#define maxn 320007
#define N 100000000
#define INF 0x3f3f3f3f
#define mod 1000000009
#define e  2.718281828459045
#define eps 1.0e18
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define ll long long
//std::ios::sync_with_stdio(false);
//cin.tie(NULL);
using namespace std;


int a[111][111];
ll row[111111];
ll col[111111];
int main()
{
    ll k=0,kk=0;
    ll n,m;
    cin>>n>>m;
    int flag=0;
    if(n<=m)
    {
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                cin>>a[i][j];
        for(int i=0; i<n; i++)
        {
            int minn=INF;
            for(int j=0; j<m; j++)
                if(minn>a[i][j])
                    minn=a[i][j];
            for(int j=0; j<m; j++)
                a[i][j]-=minn;
            for(int j=0; j<minn; j++)
            {
                row[k++]=i+1;
            }
        }
        for(int i=0; i<m; i++)
        {
            int minn=INF;
            for(int j=0; j<n; j++)
                if(minn>a[j][i])
                    minn=a[j][i];
            for(int j=0; j<n; j++)
                a[j][i]-=minn;
            for(int j=0; j<minn; j++)
            {
                col[kk++]=i+1;
            }
        }
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
            {
                if(a[i][j]!=0)
                    flag=1;
            }
        if(flag)
            cout<<"-1"<<endl;
        else
        {
            cout<<k+kk<<endl;
            for(int i=0; i<k; i++)
                cout<<"row "<<row[i]<<endl;
            for(int i=0; i<kk; i++)
                cout<<"col "<<col[i]<<endl;
        }
    }
    else
    {
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                cin>>a[i][j];
        for(int i=0; i<m; i++)
        {
            int minn=INF;
            for(int j=0; j<n; j++)
                if(minn>a[j][i])
                    minn=a[j][i];
            for(int j=0; j<n; j++)
                a[j][i]-=minn;
            for(int j=0; j<minn; j++)
            {
                col[kk++]=i+1;
            }
        }
        for(int i=0; i<n; i++)
        {
            int minn=INF;
            for(int j=0; j<m; j++)
                if(minn>a[i][j])
                    minn=a[i][j];
            for(int j=0; j<m; j++)
                a[i][j]-=minn;
            for(int j=0; j<minn; j++)
            {
                row[k++]=i+1;
            }
        }
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
            {
                if(a[i][j]!=0)
                    flag=1;
            }
        if(flag)
            cout<<"-1"<<endl;
        else
        {
            cout<<k+kk<<endl;
            for(int i=0; i<kk; i++)
                cout<<"col "<<col[i]<<endl;
            for(int i=0; i<k; i++)
                cout<<"row "<<row[i]<<endl;
        }
    }
    return 0;
}