标签：tput tip 技术分享 amp play cstring should style sar
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!
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).
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:
If there are multiple optimal solutions, output any one of them.
3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1
4
row 1
row 1
col 4
row 3
3 3
0 0 0
0 1 0
0 0 0
-1
3 3
1 1 1
1 1 1
1 1 1
3
row 1
row 2
row 3
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.
题意：给你一个n行m列的二维数组，你可以对任意行或列执行一个删除操作，即使该行或列上所有数字-1，如果能使数组全为0，则输出最小操作数以及步骤，不能则输出-1.
通过思考，有以下规则：
1.一行中最小的数字，就是能对该行执行的最大操作次数，如2 3 4 3，最多操作2次后变为0 1 2 1，此后对该行无法继续操作，对于列也是一样。
2.若对所有行执行操作，等价于对所有列执行操作
3.把对所有行或所有列的同步操作优先进行，可以达到最简化的效果
最简化操作以行列数量为准，如n=2，m=4,数组如下
k+0,k+0,k+1,k+0
k+2,k+2,k+3,k+2
若要先消除k，操作行，使数组总和-m，操作列，使数组总和-n，因此行数少，则优先列，列数少，则操作行
4.对于最简化后的数组，其任意数字等于其最大行操作数与最大列操作数之和。如上述例子
0 0 1 0
2 2 3 2
第二行为2，第三列为1，其余全为0
代码如下
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<set> using namespace std; const int MAX=1e5+10; #define INF 0x7fffffff #define ll long long #define FOR(i,n) for(i=1;i<=n;i++) int main() { int n,m,i,j,k,ans=0,a[105][105],row[105],col[105],mark1; cin>>n>>m; for(i=1;i<=n;i++) { row[i]=INF; for(j=1;j<=m;j++) { cin>>a[i][j]; row[i]=min(row[i],a[i][j]); } } for(i=1;i<=m;i++) { col[i]=INF; for(j=1;j<=n;j++) { col[i]=min(col[i],a[j][i]); } } if(n>m) { mark1=INF; for(i=1;i<=m;i++) { ans+=col[i]; mark1=min(mark1,col[i]); } for(i=1;i<=n;i++) { row[i]-=mark1; ans+=row[i]; } } else { mark1=INF; for(i=1;i<=n;i++) { ans+=row[i]; mark1=min(mark1,row[i]); } for(i=1;i<=m;i++) { col[i]-=mark1; ans+=col[i]; } } for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { if(a[i][j]!=(col[j]+row[i])) { cout<<"-1"<<endl; return 0; } } } if(n>m) { cout<<ans<<endl; for(i=1;i<=m;i++) for(j=1;j<=col[i];j++) cout<<"col "<<i<<endl; for(i=1;i<=n;i++) for(j=1;j<=row[i];j++) cout<<"row "<<i<<endl; } else { cout<<ans<<endl; for(i=1;i<=n;i++) for(j=1;j<=row[i];j++) cout<<"row "<<i<<endl; for(i=1;i<=m;i++) for(j=1;j<=col[i];j++) cout<<"col "<<i<<endl; } }
code force #419（div2）C. Karen and Game
标签：tput tip 技术分享 amp play cstring should style sar
原文地址：http://www.cnblogs.com/qq936584671/p/7043762.html
