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;

}