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 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.
题目链接：
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=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int a[550][550];
int sumr[550];
int sumc[550];
int n,m;
void buildr(int x)
{
repd(i,1,n)
{
repd(j,1,m)
{
if(i==x)
{
a[i][j]--;
}
}
}
}
void buildl(int x)
{
repd(i,1,n)
{
repd(j,1,m)
{
if(j==x)
{
a[i][j]--;
}
}
}
}
int main()
{
gg(n);
gg(m);
repd(i,1,n)
{
repd(j,1,m)
{
gg(a[i][j]);
}
}
repd(i,1,m)
{
int cnt=inf;
repd(j,1,n)
{
cnt=min(cnt,a[j][i]);
}
sumc[i]=cnt;
}
repd(i,1,n)
{
int cnt=inf;
repd(j,1,m)
{
cnt=min(cnt,a[i][j]);
}
sumr[i]=cnt;
}
int ans=0;
int flag=0;
int fu=0;
std::vector<int> r;
std::vector<int> c;
if(n<=m)
{
repd(i,1,n)
{
while(sumr[i])
{
r.pb(i);
ans++;
sumr[i]--;
buildr(i);
}
}
repd(i,1,m)
{
int cnt=inf;
repd(j,1,n)
{
cnt=min(cnt,a[j][i]);
}
sumc[i]=cnt;
}
repd(i,1,m)
{
while(sumc[i])
{
c.pb(i);
ans++;
sumc[i]--;
buildl(i);
}
}
// repd(i)
}else
{
repd(i,1,m)
{
while(sumc[i])
{
c.pb(i);
ans++;
sumc[i]--;
buildl(i);
}
}
repd(i,1,n)
{
int cnt=inf;
repd(j,1,m)
{
cnt=min(cnt,a[i][j]);
}
sumr[i]=cnt;
}
repd(i,1,n)
{
while(sumr[i])
{
r.pb(i);
ans++;
sumr[i]--;
buildr(i);
}
}
}
repd(i,1,n)
{
repd(j,1,m)
{
if(a[i][j]<0)
{
fu=min(fu,a[i][j]);
}
if(a[i][j]!=0)
{
flag=1;
break;
}
}
}
if(flag)
{
printf("-1");
return 0;
}
printf("%d\n",ans );
repd(i,0,sz(r)-1)
{
int x=r[i];
printf("row %d\n",x);
}
repd(i,0,sz(c)-1)
{
int x=c[i];
printf("col %d\n", x);
}
// db(fu);
return 0;
}
inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ‘ ‘ || ch == ‘\n‘);
if (ch == ‘-‘) {
*p = -(getchar() - ‘0‘);
while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
*p = *p * 10 - ch + ‘0‘;
}
}
else {
*p = ch - ‘0‘;
while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) {
*p = *p * 10 + ch - ‘0‘;
}
}
}
原文地址：https://www.cnblogs.com/qieqiemin/p/10323009.html