[HNOI 2014]画框

Description

\(T\) 组询问，每组询问给你个 \(2\times N\) 的带权二分图，两个权值 \(a,b\) ，让你做匹配使得 \[\sum a\times \sum b\] 最小。

\(N\leq 70,T\leq 3\)

Solution

我们用相同的思路，只是将 \(kruskal\) 求最小生成树的过程改为 \(KM\) 求最佳匹配。由于 \(KM\) 是求匹配的最大值，我们只需要将权值取相反数即可。

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 = 70, INF = 1e9;

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)); }

struct node {

    int x, y;

    node (int _x = 0, int _y = 0) {x = _x, y = _y; }

    int operator * (const node &b) const {return x*b.y-y*b.x; }

    node operator + (const node &b) const {return node(x+b.x, y+b.y); }

    node operator - (const node &b) const {return node(x-b.x, y-b.y); }

    void print() {writeln(x*y); }

}ans, A, B;

int n, a[N+5][N+5], b[N+5][N+5], c[N+5][N+5];

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]-c[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;

}

node KM() {

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

    E1[i] = -INF; E2[i] = 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; 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;

        }

    }

    }

    node ans;

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

    return ans;

}

void update(node &ans, node x) {if (x.x*x.y < ans.x*ans.y) ans = x; }

bool judge(node A, node B, node C) {return (B-A)*(C-A) >= 0; }

void doit(node A, node B) {

    int x = A.x-B.x, y = B.y-A.y; node C;

    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) c[i][j] = a[i][j]*y+b[i][j]*x;

    update(ans, C = KM());

    if (judge(A, B, C)) return;

    doit(A, C); doit(C, B);

}

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]);

    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) c[i][j] = -a[i][j];

    A = ans = KM();

    for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) c[i][j] = -b[i][j];

    update(ans, B = KM());

    doit(A, B); ans.print();

}

int main() {

    int t; read(t); while (t--) work(); return 0;

}

巴特西

[HNOI 2014]画框

Description

Solution

Code

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