[Luogu 1559]运动员最佳匹配问题

Description

求 \(2\times N\) 个点的带权二分图最佳匹配。
\(1\leq N\leq 20\)
Solution

我还是太菜了啊...到现在才学 \(KM\) 。
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('\n'))

#define lowbit(x) ((x)&(-(x)))

using namespace std;

const int N = 20, INF = ~0u>>1;

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], x;

int vis1[N+5], vis2[N+5], E1[N+5], E2[N+5], sla[N+5], match[N+5];

bool dfs(int u) {

    vis1[u] = 1;

    for (int i = 1; i <= n; i++)

    if (vis2[i] == 0) {

        int tmp = E1[u]+E2[i]-a[u][i];

        if (tmp == 0) {

        vis2[i] = 1;

        if (match[i] == 0 || dfs(match[i])) {

            match[i] = u; return true;

        }

        }else sla[i] = Min(sla[i], tmp);

    }

    return false;

}

int KM() {

    for (int i = 1; i <= n; i++) {

    E1[i] = E2[i] = match[i] = 0;

    for (int j = 1; j <= n; j++) E1[i] = Max(E1[i], a[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;

        int 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;

        }

    }

    }

    int ans = 0;

    for (int i = 1; i <= n; i++) ans += a[match[i]][i];

    return ans;

}

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(x), a[j][i] *= x;

    writeln(KM());

}

int main() {

    work(); return 0;

}