ACM HDU 2819 Swap （二分图匹配，记录过程）

Swap

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 459    Accepted Submission(s): 129
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

 

Source

2009 Multi-University Training Contest 1 - Host by TJU

 

Recommend

gaojie

 

 

#include<stdio.h>
#include<string.h>
const int MAXN=105;
int uN,vN;  //u,v数目
int g[MAXN][MAXN];//编号是0~n-1的 
int linker[MAXN];
bool used[MAXN];
int a[10000],b[10000];
bool dfs(int u)
{
    int v;
    for(v=1;v<=vN;v++)
        if(g[u][v]&&!used[v])
        {
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }    
        }  
    return false;  
}    
int hungary()
{
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=1;u<=uN;u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u))  res++;
    } 
    return res;   
}     
int main()
{
    int N;
    int i,j,u,v;
    while(scanf("%d",&N)!=EOF)
    {
        uN=vN=N;
        for(i=1;i<=uN;i++)
          for(j=1;j<=vN;j++)
            scanf("%d",&g[i][j]);
        int ans=hungary();
        //printf("%d\n",ans);
        if(ans<N){printf("-1\n");continue;}
        int res=0;
        for(i=1;i<=uN;i++)
        {
            for(j=i;j<=vN;j++)
               if(linker[j]==i) break;
            if(j!=i)
            {
                a[res]=i;b[res++]=j;
                int t=linker[i];linker[i]=linker[j];linker[j]=t;
            }    
        } 
        printf("%d\n",res);
        for(i=0;i<res;i++)
          printf("C %d %d\n",a[i],b[i]);
           
        
    } 
    return 0;   
}