C. Karen and Game

给定n行m列数字，每次可以让一行或一列都减一，求出让全部数字全为0的最小的次数，没有则输出-1；

比赛时没有考虑，n和m的大小问题，被hack了。5555555555555555555555555555

好的方法没有想到，只有一个死方法了。

  1 #include <iostream>
  2 #include <stdio.h>
  3 #define long long ll
  4 using namespace std;
  5 int a[110][110], n, m,b[110][110];
  6 bool check(){
  7     for(int i = 1; i <= n; i ++){
  8         for(int j = 1; j <= m; j ++){
  9             if(a[i][j] != 0){
 10                 return true;
 11             }
 12         }
 13     }
 14     return false;
 15 }
 16
 17 int main(){
 18     cin >> n >> m;
 19     for(int i = 1; i <= n; i ++){
 20         for(int j = 1; j <= m; j ++){
 21 //            scanf("%d",&a[i][j]);
 22             cin >> a[i][j];
 23             b[i][j] = a[i][j];
 24         }
 25     }
 26     int sum, flag = 1, i = 1,j = 1, ans = 0;
 27     if(n <= m){
 28         while(i <= n){
 29             sum = 0;
 30             flag = 1;
 31             for(int k = 1; k <= m; k ++){
 32                 if(!a[i][k]){
 33                     flag = 0;
 34                     break;
 35                 }
 36             }
 37             if(flag){
 38                 ans++;
 39                 for(int k = 1; k <= m; k ++) a[i][k] -= 1;
 40             }else i++;
 41         }
 42         while(j <= m){
 43             sum = 0;
 44             flag = 1;
 45             for(int k = 1; k <= n; k ++){
 46                 if(!a[k][j]){
 47                     flag = 0;
 48                     break;
 49                 }
 50             }
 51             if(flag){
 52                 ans ++;
 53                 for(int k = 1; k <= n; k ++) a[k][j] -= 1;
 54             }else j++;
 55         }
 56     }else{
 57         while(j <= m){
 58             sum = 0;
 59             flag = 1;
 60             for(int k = 1; k <= n; k ++){
 61                 if(!a[k][j]){
 62                     flag = 0;
 63                     break;
 64                 }
 65             }
 66             if(flag){
 67                 ans ++;
 68                 for(int k = 1; k <= n; k ++) a[k][j] -= 1;
 69             }else j++;
 70         }
 71         while(i <= n){
 72             sum = 0;
 73             flag = 1;
 74             for(int k = 1; k <= m; k ++){
 75                 if(!a[i][k]){
 76                     flag = 0;
 77                     break;
 78                 }
 79             }
 80             if(flag){
 81                 ans++;
 82                 for(int k = 1; k <= m; k ++) a[i][k] -= 1;
 83             }else i++;
 84         }
 85     }
 86     if(check()){
 87         cout << -1 << endl;
 88         return 0;
 89     }
 90     cout << ans << endl;
 91     i = j = 1;
 92     if(n <= m){
 93         while(i <= n){
 94             sum = 0;
 95             flag = 1;
 96             for(int k = 1; k <= m; k ++){
 97                 if(!b[i][k]){
 98                     flag = 0;
 99                     break;
100                 }
101             }
102             if(flag){
103                 cout << "row " << i << endl;
104                 for(int k = 1; k <= m; k ++) b[i][k] -= 1;
105             }else i++;
106         }
107         while(j <= m){
108             sum = 0;
109             flag = 1;
110             for(int k = 1; k <= n; k ++){
111                 if(!b[k][j]){
112                     flag = 0;
113                     break;
114                 }
115             }
116             if(flag){
117                 cout << "col " << j << endl;
118                 for(int k = 1; k <= n; k ++) b[k][j] -= 1;
119             }else j++;
120         }
121     }else{
122         while(j <= m){
123             sum = 0;
124             flag = 1;
125             for(int k = 1; k <= n; k ++){
126                 if(!b[k][j]){
127                     flag = 0;
128                     break;
129                 }
130             }
131             if(flag){
132                 cout << "col " << j << endl;
133                 for(int k = 1; k <= n; k ++) b[k][j] -= 1;
134             }else j++;
135         }
136         while(i <= n){
137             sum = 0;
138             flag = 1;
139             for(int k = 1; k <= m; k ++){
140                 if(!b[i][k]){
141                     flag = 0;
142                     break;
143                 }
144             }
145             if(flag){
146                 cout << "row " << i << endl;
147                 for(int k = 1; k <= m; k ++) b[i][k] -= 1;
148             }else i++;
149         }
150     }
151     return 0;
152 }