hdu 2819 Swap

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”.

//在N*N的方格中选 N个 行列不同的点、 然后把这些点排个顺序
// 在N*N的方格中选 N个 行列不同的点 -->明显的二分图匹配模型 
#include <iostream>
#include <vector>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N=110;
bool visit[N];
int match[N];
vector <int> G[N];
int rc[N][2];
bool dfs(int u)
{
    int i,v;
    for(i=0;i<G[u].size();i++)
     {
         v=G[u][i];
         if(!visit[v])
         {
              visit[v]=true;
              if(!match[v]||dfs(match[v]))
              {
                  match[v]=u;
                  return true;
              }
         }
     }
     return false;
}
int main()
{
   int n,m;
    while(scanf("%d",&n)!=EOF)
    {
        int i,j;
        for(i=1;i<=n;i++)
         for(G[i].clear(),j=1;j<=n;j++)
          {
              scanf("%d",&m);
              if(m) G[i].push_back(j);
          }
          int ans=0;
          memset(match,0,sizeof(match));
          for(i=1;i<=n;i++)
          {
              memset(visit,0,sizeof(visit));
              if(dfs(i))   ans++;
          }
          if(ans!=n) { printf("-1\n");continue;}
          m=0;
          for(i=1;i<n;i++)
            if(match[i]!=i)
            {
                for(j=i+1;j<=n;j++)
                   if(match[j]==i)
                   {   m++;
                      rc[m][0]=match[j];
                      rc[m][1]=match[i];
                      match[j]=match[i];
                   }
            }
          printf("%d\n",m);
          for(i=1;i<=m;i++)
            printf("R %d %d\n",rc[i][0],rc[i][1]);
    }

    return 0;
}