hdu2819二分图匹配

E - Swap

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 2819

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 
       
       
       
       
 
       
       
       
       

       
       
       
       
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
using namespace std;
const int maxn=105;
int g[maxn][maxn];
int linker[maxn];
bool used[maxn];
int n;

bool dfs(int u)
{
    for(int i=1; i<=n; i++)
    {
        if(!used[i]&&g[u][i])
        {
            used[i]=true;
            if(!linker[i]||dfs(linker[i]))
            {
                linker[i]=u;
                return true;
            }
        }
    }
    return false;
}
int solve()
{
    int ans=0;
    memset(linker,0,sizeof(linker));
    for(int i=1; i<=n; i++)
    {
        memset(used,false,sizeof(used));
        if(dfs(i))
            ans++;
    }
    return ans;
}
int main()
{
    int a[maxn],b[maxn];
    int op;
    while(scanf("%d",&n)!=-1)
    {
        memset(g,0,sizeof(g));
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                  scanf("%d",&op);
                  if(op)
                    g[i][j]=1;
            }

        int temp=solve();
        if(temp!=n)
        {
            cout<<-1<<endl;
            continue;
        }
        int tot = 0,j;

        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            if(i != j&&linker[j]==i)//交换第i列和第j列
            {
                a[tot] = i;
                b[tot] = j;
                tot ++;//记录结果
                int t = linker[i];
                linker[i] = linker[j];
                linker[j] = t;
            }
        }
        cout<<tot<<endl;
        for(int i=0; i<tot; i++)
        {
            printf("C %d %d\n",a[i],b[i]);
        }
    }
    return 0;
}