luogu P4100 [HEOI2013]钙铁锌硒维生素

https://www.luogu.com.cn/problem/P4100
题目大意：

首先构造两个矩阵A，B
把A,B转置一下变成$A^T,B^T$
就相当于是写成把若干个向量拼成矩阵的形式
然后可以得到
$$B=RA$$保证$R[i][j]!=0$才能让$A[i]$用$B[j]$代替，不然说明$B[j]$会与除了$A[i]$外的向量线性相关
所以$$R=B{A}^{-1}$$
具体做的话可以在$A$后面接一个伴随矩阵$B$然后把$A$消元成单位矩阵，这时候$B$就变成$R$了
再根据$R$连边跑最大匹配(字典序最小的)即可
code:

#include<bits/stdc++.h>
#define N 605
using namespace std;
const double eps = 1e-9;
int vis[N], match[N], n;
double a[N][N], b[N][N];
int dfs(int u) {
    for(int v = 1; v <= n; v ++) if(!vis[v] && fabs(b[u][v]) > eps) {
        vis[v] = 1;
        if(!match[v] || dfs(match[v])) {
            match[v] = u;
            return 1;
        }
    }
    return 0;
}
int dfss(int u, int fr) {
    for(int v = 1; v <= n; v ++) if(!vis[v] && fabs(b[u][v]) > eps) {
        vis[v] = 1;
        if(match[v] == fr || (match[v] > fr && dfss(match[v], fr))) {
            match[v] = u;
            return v;
        }
    }
    return 0;
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= n; j ++)
            scanf("%lf", &a[j][i]);
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= n; j ++)
            scanf("%lf", &b[j][i]);
    for(int i = 1; i <= n; i ++) {
        int j = i;
        for(int k = i + 1; k <= n; k ++) if(fabs(a[k][i]) > fabs(a[j][i])) j = k;
        if(j != i) swap(a[i], a[j]), swap(b[i], b[j]);
        double x = a[i][i];
        if(fabs(x) < eps) { printf("NIE"); return 0;}
        for(int j = 1; j <= n; j ++) a[i][j] /= x, b[i][j] /= x;
        for(int j = 1; j <= n; j ++) if(j != i) {
            double t = a[j][i];
            for(int k = 1; k <= n; k ++) {
                a[j][k] -= a[i][k] * t;
                b[j][k] -= b[i][k] * t;
            }
        }
    }
//    for(int i = 1; i <= n; i ++) {
//        for(int j = 1; j <= n; j ++) printf("%.3lf ", b[i][j]); printf("\n");
//    }
    for(int i = 1; i <= n; i ++) {
        memset(vis, 0, sizeof vis);
        if(!dfs(i)) { printf("NIE"); return 0;}
    }
  //  for(int i = 1; i <= n; i ++) printf(" %d ", match[i]); printf("\n");
    puts("TAK");
    for(int i = 1; i <= n; i ++) {
        memset(vis, 0, sizeof vis);
        printf("%d\n", dfss(i, i));
    }
    return 0;
}