CSA：Flipping Matrix（二分图匹配 & 思维）

Flipping Matrix

Time limit:  1000 ms
Memory limit:  256 MB

You are given a binary matrix $A$ of size $N\times N$. You are allowed to perform the following two operations:

• Take two rows and swap them. If we want to swap rows $x$ and $y$, we'll encode this operation as R x y.
• Take two columns and swap them. If we want to swap columns $x$ and $y$, we'll encode this operation as C x y.

Is it possible to obtain only values of $1$ on the main diagonal of $A$ by performing a sequence of at most $N$ operations? If so, print the required operations.

Standard input

The first line contains $N$.

The next $N$ lines contain $N$ binary values separated by spaces, representing $A$.

Standard output

If there is no solution, print $-1$.

Otherwise, print every operation on a separated line.

Constraints and notes

• 2 \leq N \leq 10^32≤N≤10​3​​ 
• 0 \leq A_{i, j} \leq 10≤A​i,j​​≤1 for every $1\le i,j\le N$

Input	Output
3 0 0 1 0 1 0 1 0 0	C 1 3
4 1 1 0 0 0 1 0 1 1 1 0 0 0 0 0 1	-1
5 0 1 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 1 1 0 1 0 0 0 0	R 1 5 C 2 3

题意：给一个01矩阵，问能否通过若干次行或列的交换，使得主对角线上全为1，输出其中一种交换方案。

思路：如果有解，一定能仅用行交换或者仅用列交换完成，因为无论怎么交换，同一行的1永远在同一行，最终对角线上的1一定来自不同的行和不同的列，假设去除其他无用的1，交换行和交换列效果等价。我们对交换列讨论，如果a[i][j]为1，i到j建边，表示j列换到i列可以匹配到一个1，最后跑一次最大匹配看是否为N即可。

# include 
using namespace std;
const int maxn = 1e3+3;
int a[maxn][maxn];
int l[maxn], vis[maxn];
vectorg[maxn];
int dfs(int u)
{
    for(int v:g[u])
    {
        if(!vis[v])
        {
            vis[v]=1;
            if(l[v]==0 || dfs(l[v]))
            {
                l[v] = u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int n, ans=0;
    scanf("%d",&n);
    for(int i=1; i<=n; ++i)
        for(int j=1; j<=n; ++j) scanf("%d",&a[i][j]);
    for(int i=1; i<=n; ++i)
        for(int j=1; j<=n; ++j) if(a[i][j]) g[i].push_back(j);
    for(int i=1; i<=n; ++i)
    {
        memset(vis, 0, sizeof(vis));
        if(dfs(i)) ++ans;
    }
    if(ans < n) puts("-1");
    else
    {
        memset(vis, 0, sizeof(vis));
        for(int i=1; i<=n; ++i)
        {
            if(vis[i]) continue;
            int j=i;
            while(l[j]!=i)
            {
                vis[l[j]]=1;
                printf("C %d %d\n",i,l[j]);
                j=l[j];
            }
        }
    }
    return 0;
}