HDU2819:Swap

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Swap Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1046    Accepted Submission(s): 348 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

=====================================题目大意=====================================

给出一个N阶的01方阵，求解通过交换行和交换列使得方阵的主对角线元素全部为1的任意一种方案。

=====================================算法分析=====================================

在给定的方阵中选取N个两两之间不同行不同列的1，然后便一定可以通过只交换行（或只交换列）移动这些1以得到所需的方阵。

显然二分图的最大匹配：将行和列作为两个点集，矩阵中的元素1作为连接其所在行和所在列的边，使用匈牙利算法求解即可。

=======================================代码=======================================

#include<stdio.h>
#include<string.h>

int N,Linker[105],RecordSwap[105][2];

bool Edge[105][105],Vis[105];

void Swap(int &A,int &B)
{
	int tmp=A;  A=B;  B=tmp;
}

bool DFS(int U)
{
	for(int v=1;v<=N;++v)
	{
		if(Edge[U][v]&&!Vis[v])
        {
			Vis[v]=true;
			if(Linker[v]==-1||DFS(Linker[v]))
			{
				Linker[v]=U;
				return true;
			}
         }
     }
     return false;
}

int Hungary()
{
    memset(Linker,-1,sizeof(Linker));
	int ans=0;
    for(int u=1;u<=N;++u)
    {
        memset(Vis,0,sizeof(Vis));
        if(DFS(u)) { ++ans; }
    } 
    return ans;  
}

int main()
{
	while(scanf("%d",&N)==1)
	{
		for(int i=1;i<=N;++i)
		{
			for(int j=1;j<=N;++j)
			{
				scanf("%d",&Edge[i][j]);
			}
		}
		if(Hungary()<N)
		{
			printf("-1\n");  continue;
		}
		int sum=0;
		for(int i=1;i<=N;++i) if(Linker[i]!=i)
		{
			for(int j=i+1;j<=N;++j) if(Linker[j]==i)
			{
				++sum;
				RecordSwap[sum][0]=i;
				RecordSwap[sum][1]=j;
				Swap(Linker[i],Linker[j]);
			}
		} 
		printf("%d\n",sum);
        for(int i=1;i<=sum;i++)
		{
            printf("C %d %d\n",RecordSwap[i][0],RecordSwap[i][1]);
		}
    }
	return 0;
}