Swap HDU - 2819（二分图最大匹配）-优快云博客

本文探讨了一种算法问题，如何通过在给定的01矩阵中仅交换行或列，使得主对角线的所有元素变为1。策略涉及最大匹配和行列交换操作，展示了解决思路和示例输入输出。

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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

题意

给一个大小为n*n的01矩阵，问能否通过交换行或列，使得主对角线上的数全为1？

思路

先求出最大匹配数，如果等于n，就可以实现。

只换行或者只换列，就能使对角线上的数全为1。

枚举每一行，对于第i行，找到与第i列匹配的那一行j，然后交换第i行和第j行。这样就满足题目要求

代码

#include<bits/stdc++.h>
using namespace std;
int a[110][110];
int book[110];
int x[110],y[110];
int n;
int dfs(int u)
{
    for(int i=1; i<=n; i++)
    {
        if(book[i]==0&&a[u][i]==1)
        {
            book[i]=1;
            if(y[i]==0||dfs(y[i]))
            {
                y[i]=u;
                x[u]=i;
                return 1;
            }
        }
    }
    return 0;
}
int hungary()
{
    int ans=0;
    memset(x,0,sizeof(x));
    memset(y,0,sizeof(y));
    for(int i=1; i<=n; i++)
    {
        memset(book,0,sizeof(book));
        if(dfs(i))
            ans++;
    }
    return ans;
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                scanf("%d",&a[i][j]);
        int cnt=hungary();
        //printf("%d\n",cnt);
         if(cnt<n)
             printf("-1\n");
         else
         {
             int op[110][2];
             int sum=0;
             for(int i=1;i<n;i++)
             {
                 for(int j=i+1;j<=n;j++)
                 {
                     if(x[j]==i)
                     {
                         swap(x[j],x[i]);
                         op[++sum][0]=i;
                         op[sum][1]=j;
                         break;
                     }
                 }
             }
             printf("%d\n",sum);
             for(int i=1;i<=sum;i++)
                 printf("R %d %d\n",op[i][0],op[i][1]);
         }
    }
    return 0;
}