HDU2819 Swap

Swap

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4036    Accepted Submission(s): 1493
Special Judge

Problem Description

Given an N*N matrix with each entry equal to 0 or 1. You can swap any two rows or any two columns. Can you find a way to make all the diagonal entries equal to 1?

 

Input

There are several test cases in the input. The first line of each test case is an integer N (1 <= N <= 100). Then N lines follow, each contains N numbers (0 or 1), separating by space, indicating the N*N matrix.

 

Output

For each test case, the first line contain the number of swaps M. Then M lines follow, whose format is “R a b” or “C a b”, indicating swapping the row a and row b, or swapping the column a and column b. (1 <= a, b <= N). Any correct answer will be accepted, but M should be more than 1000.

If it is impossible to make all the diagonal entries equal to 1, output only one one containing “-1”.

 

Sample Input

2
0 1
1 0
2
1 0
1 0

 

Sample Output

1
R 1 2
-1

 

题意：给你一个由0和1组成的图让你通过交换行或者列使得这张图主对角线都为1，不过不可以的话输出-1，可以的话输出交换几次，并且输出交换的方法。（因为交换方法很多，输出一组即可）

理性地想一下可以的情况，每行每列都要有个1，任意两行或者两列不能出现多个1。并且通过行交换出来的图，用列或者行列并用也可以交换出来。

至于为什么要用二分图，我认为用二分图一是好判断是不是可以变成主对角线为1的图，二是可以写出交换方法。

我们把行当做一个图，1所在的列当做另一个图，这样一来1所在的行列也就是1的坐标就是连线方式，x代表行，y代表列。当最大匹配等于n时就代表可以换成要求的图。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
int n;
int mapp[105][105],used[105],pp[105];
int st[105],ed[105];
int found(int x)
{
    for(int i=1;i<=n;i++)
    {
        if(mapp[x][i]&&!used[i])
        {
            used[i]=1;
            if(!pp[i]||found(pp[i]))
            {
                pp[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int aa()
{
    int sum=0;
    memset(pp,0,sizeof(pp));
    for(int i=1;i<=n;i++)
    {
        memset(used,0,sizeof(used));
        if(found(i))sum++;
    }
    return sum;
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
            scanf("%d",&mapp[i][j]);
        int ans=aa();
        if(ans!=n)
        {
            printf("-1\n");
        }
        else
        {
            int ans=0;
            for(int i=1;i<=n;i++)//从行找起
            {
                if(i!=pp[i])//如果行号不对应列号
                {
                    for(int j=1;j<=n;j++)//去列号里边找
                    {
                        if(i==pp[j])//从列里找到相同的
                        {
                            st[ans]=i;//记录
                            ed[ans]=j;
                            ans++;
                            swap(pp[i],pp[j]);//交换列
                            break;
                        }
                    }
                }
            }
            printf("%d\n",ans);
            for(int i=0;i<ans;i++)
            {
                printf("C %d %d\n",st[i],ed[i]);
            }
        }
    }
}