Hdu 2819 Swap 匈牙利算法

题目：

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

思路：

将这个方阵抽象成一个邻接矩阵，0表示不存在边，1表示存在边，题目的目标状态是对角线元素为1，所以表明，每一横行和每一竖行都必须有边，否则不能转换成目标状态。

此题可以通过匈牙利算法来实现匹配，然后遍历来实现交换元素。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=105;
int n;
int a[maxn][maxn];
int match[maxn];
int vis[maxn];
bool Find(int x)
{
    for (int i=1;i<=n;i++)
    {
        if(a[x][i]&&!vis[i])
        {
            vis[i]=1;
            if(match[i]==-1||Find(match[i]))
            {
                match[i]=x;
                return true;
            }
        }
    }
    return false;
}
int alor()
{
    int sum=0;
    for (int i=1;i<=n;i++)
    {
        memset (vis,0,sizeof(vis));
        if(Find(i)) sum++;
    }
    return sum;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset (match,-1,sizeof(match));
        for (int i=1;i<=n;i++)
        {
            for (int j=1;j<=n;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        if(alor()!=n)
        {
            printf("-1\n");
            continue;
        }
        else
        {
            int cnt=0;
            int opa[maxn*maxn],opb[maxn*maxn];
            for (int i=1;i<=n;i++)
            {
                if(match[i]==i) continue;
                for (int j=i+1;j<=n;j++)
                {
                    if(match[j]==i)
                    {
                        opa[cnt]=i,opb[cnt]=j;
                        swap(match[i],match[j]);
                        cnt++;
                        break;
                    }
                }
            }
            printf("%d\n",cnt);
            for (int i=0;i<cnt;i++) printf("C %d %d\n",opa[i],opb[i]);
        }
    }
    return 0;
}