Karen and Game CodeForces - 816C （暴力+构造）

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 mcolumns. 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 g_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 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 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 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.

题目链接：

CodeForces - 816C

题意：

给定一个n*m的矩阵，

每一个操作你可以选择一行或者一列，使该行或列的数值全部加1，求使用最小的操作次数使一个全部为0的矩阵变成给定矩阵，

如果不可能实现请输出-1.

思路：

预处理出每一行和每一列的最小值，

然后以n和m的关系进行分类处理，

如果n<=m，就先处理行再处理列，这样可以用最小的次数，

反而反之。

举例：

1 1 1 1

1 1 1 1

1 1 1 1

如果先处理列，就要4次，处理行只需要3次。

接下来：

如果每一行或列的最小值大于0，那么我们就可以处理最小值次，然后每一次处理，暴力的把数组中的对应元素减去1，

行和列全部处理好后，去n*m扫一边数组，如果还有数大于0，那么就是无法实现的情况。

每一步具体为什么可能需要大家自己好好思考。

细节见我的代码。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=;while(b){if(b%)ans=ans*a%MOD;a=a*a%MOD;b/=;}return ans;}
inline void getInt(int* p);
const int maxn=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int a[][];
int sumr[];
int sumc[];
int n,m;
void buildr(int x)
{
    repd(i,,n)
    {
        repd(j,,m)
        {
            if(i==x)
            {
                a[i][j]--;
            }

        }
    }

}
void buildl(int x)
{
    repd(i,,n)
    {
        repd(j,,m)
        {
            if(j==x)
            {
                a[i][j]--;
            }
        }
    }
}
int main()
{
    gg(n);
    gg(m);
    repd(i,,n)
    {
        repd(j,,m)
        {
            gg(a[i][j]);

        }
    }
    repd(i,,m)
    {
        int cnt=inf;
        repd(j,,n)
        {
            cnt=min(cnt,a[j][i]);
        }
        sumc[i]=cnt;
    }
    repd(i,,n)
    {
        int cnt=inf;
        repd(j,,m)
        {
            cnt=min(cnt,a[i][j]);
        }
        sumr[i]=cnt;
    }
    int ans=;
    int flag=;
    int fu=;
    std::vector<int> r;
    std::vector<int> c;
    if(n<=m)
    {
        repd(i,,n)
        {
            while(sumr[i])
            {
                r.pb(i);
                ans++;
                sumr[i]--;
                buildr(i);
            }
        }
        repd(i,,m)
        {
            int cnt=inf;
            repd(j,,n)
            {
                cnt=min(cnt,a[j][i]);
            }
            sumc[i]=cnt;
        }
        repd(i,,m)
        {
            while(sumc[i])
            {
                c.pb(i);
                ans++;
                sumc[i]--;
                buildl(i);
            }
        }
//        repd(i)
    }else
    {
        repd(i,,m)
        {
            while(sumc[i])
            {
                c.pb(i);
                ans++;
                sumc[i]--;
                buildl(i);
            }
        }
        repd(i,,n)
        {
            int cnt=inf;
            repd(j,,m)
            {
                cnt=min(cnt,a[i][j]);
            }
            sumr[i]=cnt;
        }
        repd(i,,n)
        {
            while(sumr[i])
            {
                r.pb(i);
                ans++;
                sumr[i]--;
                buildr(i);
            }
        }
    }
    repd(i,,n)
    {
        repd(j,,m)
        {
            if(a[i][j]<)
            {
                fu=min(fu,a[i][j]);
            }
            if(a[i][j]!=)
            {
                flag=;
                break;
            }
        }
    }
    if(flag)
    {

        printf("-1");
        return ;
    }
    printf("%d\n",ans );
    repd(i,,sz(r)-)
    {
        int x=r[i];
        printf("row %d\n",x);
    }
    repd(i,,sz(c)-)
    {
        int x=c[i];
        printf("col %d\n", x);
    }
//    db(fu);
    return ;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '');
        while ((ch = getchar()) >= '' && ch <= '') {
            *p = *p *  - ch + '';
        }
    }
    else {
        *p = ch - '';
        while ((ch = getchar()) >= '' && ch <= '') {
            *p = *p *  + ch - '';
        }
    }
}