标签：create contains wing int creat printf 图片 beat original
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:
If there are multiple optimal solutions, output any one of them.
Sample Input
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
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.
分析这种棋盘的特性，要加加一行，要加加一列，且不能减，且棋盘初始都为零。所以如果同一行上出现了不同的数，则数大的一列必然有该列的加，行同理。
所以可以先处理列，把所有列加过的都还原回去，所以只剩下了由个别的行改变所导致的棋盘，再用同样的方法将行还原回去，最后如果棋盘仍然不为零的话，说明所有的行都进行过相同次数的
加（这里要注意！！！如果最后棋盘不为零的话，既有可能是所有的行进行了操作，也有可能是所有的列进行了操作，因为题目让求最少的操作次数，所以应判断行和列谁小操作的谁）。
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<string> 5 #include<cmath> 6 #include<algorithm> 7 #include<queue> 8 #include<stack> 9 #include<deque> 10 #include<map> 11 #include<iostream> 12 using namespace std; 13 typedef long long LL; 14 const double pi=acos(-1.0); 15 const double e=exp(1); 16 const int N = 100009; 17 18 LL con[110][110]; 19 LL check[110][110]; 20 LL col[110],row[110]; 21 22 int main() 23 { 24 LL i,p,j,m,n; 25 LL flag=0; 26 scanf("%lld%lld",&m,&n); 27 LL head=10000000,tail=10000000,mid; 28 29 for(i=1; i<=m; i++) 30 { 31 for(j=1; j<=n; j++) 32 { 33 scanf("%lld",&con[i][j]); 34 if(con[1][j]<head) 35 head=con[1][j]; 36 if(con[i][1]<tail) 37 tail=con[i][1]; 38 } 39 } 40 41 if(m<=n) 42 { 43 for(i=1; i<=n; i++) 44 { 45 if(con[1][i]>head) 46 col[i]+=con[1][i]-head; 47 } 48 mid=tail-col[1]; 49 for(i=1; i<=m; i++) 50 { 51 if(con[i][1]>tail) 52 row[i]+=con[i][1]-tail; 53 row[i]+=mid; 54 } 55 } 56 else 57 { 58 for(i=1; i<=m; i++) 59 { 60 if(con[i][1]>tail) 61 row[i]+=con[i][1]-tail; 62 } 63 mid=head-row[1]; 64 for(i=1; i<=n; i++) 65 { 66 if(con[1][i]>head) 67 col[i]+=con[1][i]-head; 68 col[i]+=mid; 69 } 70 } 71 72 flag=0; 73 74 for(i=1; i<=n; i++) 75 { 76 if(col[i]) 77 { 78 flag+=col[i]; 79 for(j=1; j<=m; j++) 80 check[j][i]+=col[i]; 81 } 82 } 83 for(i=1; i<=m; i++) 84 { 85 if(row[i]) 86 { 87 flag+=row[i]; 88 for(j=1; j<=n; j++) 89 check[i][j]+=row[i]; 90 } 91 } 92 93 for(i=1; i<=m; i++) 94 { 95 for(j=1; j<=n; j++) 96 { 97 if(con[i][j]!=check[i][j]) 98 break; 99 } 100 if(j<=n) 101 { 102 flag=-1; 103 break; 104 } 105 } 106 107 108 109 if(flag==-1) 110 printf("-1\n"); 111 else 112 { 113 printf("%lld\n",flag); 114 for(i=1; i<=n; i++) 115 { 116 while(col[i]) 117 { 118 col[i]--; 119 printf("col %lld\n",i); 120 } 121 } 122 for(i=1; i<=m; i++) 123 { 124 while(row[i]) 125 { 126 row[i]--; 127 printf("row %lld\n",i); 128 } 129 } 130 } 131 return 0; 132 }
标签：create contains wing int creat printf 图片 beat original
原文地址：https://www.cnblogs.com/daybreaking/p/9746733.html
