0/1分数规划
二分答案 + 费用流

#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>

#define INF 1e9 + 7
#define eps 1e-7

using namespace std;

const int Maxn = 205;
const int N = 20100;

struct node {
    double val; short cap, to, next;    
} e[N];

short n, a[Maxn][Maxn], b[Maxn][Maxn], tot = 1, h[Maxn], S, T, s, pre[Maxn], cur[Maxn];
double dis[Maxn];
bool vis[Maxn];

inline char NC(void)
{
  static char buf[100000], *p1 = buf, *p2 = buf;
  if (p1 == p2) {
    p2 = (p1 = buf) + fread(buf, 1, 100000, stdin);
    if (p1 == p2) return EOF;
  }
  return *p1++;
}

inline int read() {
    static char c; c = NC();  int x;
    static int minus; minus = 1;
    for (x = 0; !(c >= '0' && c <= '9'); c = NC()) if(c == '-') minus = -1;
    for (; c >= '0' && c <= '9'; x = x * 10 + c - '0', c = NC()); x *= minus;
    return x;
}

inline void Link(int u, int v, int c, double w) {
    e[++tot].to = v; e[tot].next = h[u];
    e[tot].cap = c; e[tot].val = w; h[u] = tot;    
}

inline void Add(int u, int v, int c, double w) {
    Link(u, v, c, w); Link(v, u, 0, -w);
}

short Dfs(short u, short Min) {
    if(u == T) return Min;
    int w, used = 0; vis[u] = 1;
    for(short i = h[u]; i; i = e[i].next){
        if(!vis[e[i].to] && e[i].cap && dis[e[i].to] + e[i].val == dis[u]){
            w = Dfs(e[i].to, min(e[i].cap, Min));
            e[i].cap -= w; e[i ^ 1].cap += w;
            used += w; if(used == Min)return Min;
        }
    }
    if(used == 0) vis[u] = 1;
    return used;
}

inline bool Bfs() {
    double Min = 1e100;
    for(short i = S; i <= T; ++i)
        if(vis[i])
            for(short j = h[i]; j; j = e[j].next)
                if(!vis[e[j].to] && e[j].cap){
                    Min = min(Min, dis[e[j].to] + e[j].val - dis[i]);
                }
    if(Min == 1e100) return 0;
    for(short i = S; i <= T; ++i) if(vis[i]) {dis[i] += Min; vis[i] = 0;}
    return 1;
}

double Check1(float c) {
    memset(dis, 0 ,sizeof(dis));
    for(register short i = 2; i <= tot; ++i) {
        if(e[i].to >= n + 1 && e[i].to <= n + n && e[i ^ 1].to >= 1 && e[i ^ 1].to <= n) {
            int u = e[i ^ 1].to, v = e[i].to - n;
            e[i].cap = 1; e[i].val = c * (double)b[u][v] - (double)a[u][v];
            e[i ^ 1].cap = 0; e[i ^ 1].val = -e[i].val;
        }
        if(e[i ^ 1].to == S) e[i].cap = 1, e[i ^ 1].cap = 0;
        if(e[i].to == T) e[i].cap = 1, e[i ^ 1].cap = 0;
    }
    double res = 0; int s = 0;
    do{
        double temp = 0;
        while(temp = Dfs(S, 32000)){
            memset(vis, 0, sizeof(vis));
            res += temp * dis[S];
        }
    } while(Bfs());    
    
    double A = 0, B = 0;
    for(int i = 2; i <= tot; ++i) {
        int u = e[i ^ 1].to, v = e[i].to;
        if(u >= 1 && u <= n && v >= n + 1 && v <= n + n && e[i].cap == 0) {
            A += a[u][v - n]; B += b[u][v - n];
        }
    }
    return A / B;
}

int main() {
//    freopen("ball.in", "r", stdin);
//    freopen("ball.out", "w", stdout);
    n = read(); T = n + n + 1;
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= n; ++j) a[i][j] = read();
    }
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= n; ++j) b[i][j] = read();
    }
    for(int i = 1; i <= n; ++i) {
        Add(S, i, 1, 0); Add(i + n, T, 1, 0);
        for(int j = 1; j <= n; ++j) Add(i, j + n, 1, 0);
    }
    double l = 0, r;
    while(true) {
        r = Check1(l);
        if(r - l < eps) break;
        l = r;
    }
    printf("%.6lf\n", r);
    return 0;
}