G - Lines of Containers

A shipment of Nlogs, the main export product from Nlogonia, is in the harbour, in containers, ready
to be shipped. All containers have the same dimensions and all of them are cubes. Containers are
organized in the cargo terminal in L lines and C columns, for a total of LC containers. Each container
is marked with a distinct identification number, from 1 to LC.
Each one of the L container lines will be loaded in a different ship. To make it simpler when
unloading at each destination country, containers from a line must be organized such that the identifiers
are in sequential order. More precisely, the first line must have the containers identified from 1 to C in
increasing order, line 2 must have containers numbered from C + 1 to 2C (in increasing order), and so
on until the last line. which will have containers numbered (L − 1)C + 1 to LC (again, in increasing
order). Figure (a) shows the organization of a shipment with 5 lines and 4 columns of containers.
A crane is able to move either a full line or a full column of containers. It cannot move other
groupments or individual containers.
In night before the loading, a group of workers operated the cranes to swap shipment lines and
columns as a way of protest because of low salaries. Figure (b) shows the configuration after swapping
lines 1 and 4; Figure (c) shows the configuration after another swap, this time between colummns 2
and 3.
The loading must be done today, but before starting the containers must be reorganized as described
previously. You must write a program which, given the information about the position of every container
after the protest, determines whether you can reposition the containers in such way that every one of
them is in its expected positions, using only cranes. If repositioning is possible, you must also calculate
the smallest number of column and line swaps needed to do it.
Input
The input contains several test cases. The first line of of a test case contains two integers L and C
which describe, respectively, the number of lines and columns of the shipment. The next L lines show
the configuration of the containers after the workers protest. Each of these L lines will have C integers
Xl,c , indicating the position of a container. Every integer between 1 and LC appears exactly once in
the input.
Output
For each test case your program must produce a single line, containing a single integer, the minimum
number of column and line swaps needed to place the containers in their original positions. If there isUniversidad de Valladolid OJ: 12652 – Lines of Containers 2/2
no way to place the containers in their original positions using only cranes, the line must contain only
the character ‘*’.
Restrictions
• 1 ≤ L ≤ 300
• 1 ≤ C ≤ 300
• 1 ≤ Xl,c ≤ LC
Sample Input
2 2
3 4
1 2
3 3
9 2 4
5 8 7
6 1 3
5 4
13 15 14 16
5 7 6 8
9 11 10 12
1 3 2 4
17 19 18 20
Sample Output
1
*
2

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    int num[305][305];
    int n,m;
    int a=0,b=0;
    int sum=0;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            cin>>num[i][j];
            if(num[i][j]==1){
                a=i;
                b=j;
            }
        }
    }
    for(int i=1;i<=m;i++){
        if(num[a][i]!=i){
            int flag=0;
            for(int j=i+1;j<=m;j++){
                if(num[a][j]==i){
                    if(i==1) b=j;
                    sum++;
                    flag=1;
                    for(int t=1;t<=n;t++){
                          swap(num[t][j],num[t][i]);
                    }
                }
            }
            if(flag==0){
                cout<<'*'<<endl;
                return 0;
            }
        }
    }
    for(int i=1;i<=n;i++){
        if(num[i][b]!=(i-1)*m+b){
            int flag=0;
            for(int j=i+1;j<=n;j++){
                if(num[j][b]==(i-1)*m+b){
                    sum++;
                    flag=1;
                    for(int t=1;t<=m;t++){
                          swap(num[j][t],num[i][t]);
                    }
                }
            }
             if(flag==0){
                cout<<'*'<<endl;
                return 0;
            }
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(num[i][j]!=((i-1)*m+j)){
                cout<<'*'<<endl;
                return 0;
            }
        }
    }
    cout<<sum<<endl;
    return 0;
}