[SDOI 2017]新生舞会

Description

题库链接

给你个 (2 imes N) 的带权二分图，两个权值 (a,b) ，让你做匹配使得 [frac{sum a}{sum b}] 最大。

(1leq Nleq 100)

Solution

依旧是 (01) 分数规划的套路。我们二分答案 (mid) ，将每条边的边权修改为 (a-midcdot b) 。再跑一边最佳匹配看答案是否 (geq 0) 。若满足，则左端点右移，不满足就右端点左移。记得边权可能为负，所以初始化左标杆时不能默认最小值为 (0) 。

Code

//It is made by Awson on 2018.3.8
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('
'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 100;
const double eps = 1e-7, INF = 1e99;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, a[N+5][N+5], b[N+5][N+5]; double c[N+5][N+5];
double E1[N+5], E2[N+5], sla[N+5]; int vis1[N+5], vis2[N+5], match[N+5];

bool dfs(int u) {
    vis1[u] = 1;
    for (int i = 1; i <= n; i++)
    if (vis2[i] == 0) {
        double tmp = E1[u]+E2[i]-c[u][i];
        if (fabs(tmp) < eps) {
        vis2[i] = 1;
        if (match[i] == 0 || dfs(match[i])) {
            match[i] = u; return true;
        }
        }else sla[i] = min(sla[i], tmp);
    }
    return false;
}
bool KM() {
    for (int i = 1; i <= n; i++) {
    E1[i] = -INF, E2[i] = 0, match[i] = 0;
    for (int j = 1; j <= n; j++) E1[i] = max(E1[i], c[i][j]);
    }
    for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) sla[j] = INF;
    while (1) {
        for (int j = 1; j <= n; j++) vis1[j] = vis2[j] = 0;
        if (dfs(i)) break;
        double tmp = INF;
        for (int j = 1; j <= n; j++) if (vis2[j] == 0) tmp = min(tmp, sla[j]);
        for (int j = 1; j <= n; j++) {
        if (vis1[j]) E1[j] -= tmp;
        if (vis2[j]) E2[j] += tmp; else sla[j] -= tmp;
        }
    }
    }
    double ans = 0.; for (int i = 1; i <= n; i++) ans += c[match[i]][i];
    return ans >= 0.;
}
void work() {
    read(n);
    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) read(a[i][j]);
    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) read(b[i][j]);
    double L = 0, R = 1e6;
    while (R-L > eps) {
    double mid = (R+L)/2.;
    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) c[i][j] = a[i][j]-mid*b[i][j];
    if (KM()) L = mid; else R = mid;
    }
    printf("%.6lf
", (L+R)/2.);
}
int main() {
    work(); return 0;
}