【LG3236】[HNOI2014]画框

【LG3236】[HNOI2014]画框

题面

洛谷

题解

和这题一模一样。

将最小生成树换成\(KM\)即可。

关于复杂度，因为决策点肯定在凸包上，且\(n\)凸包的期望点数为\(\sqrt {\ln n}\)

所以\(n!\)个点的期望点数为\(\sqrt {\ln n!}=\sqrt {\sum_{i=1}^ni}\)

所以总复杂度\(O(\sqrt {\ln n!}*n^4)\)

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int INF = 1e9;
const int MAX_N = 100;
struct Point { int x, y; } ;
Point operator - (const Point &l, const Point &r) { return (Point){l.x - r.x, l.y - r.y}; }
int cross(const Point &l, const Point &r) { return l.x * r.y - l.y * r.x; }
int N, ans;
namespace Gra {
	int A[MAX_N][MAX_N], B[MAX_N][MAX_N];
    int g[MAX_N][MAX_N], lx[MAX_N], ly[MAX_N], sla[MAX_N], match[MAX_N];
    bool visx[MAX_N], visy[MAX_N];
    void build(int wx, int wy) {
        for (int i = 1; i <= N; i++)
            for (int j = 1; j <= N; j++)
                g[i][j] = -(wx * A[i][j] + wy * B[i][j]);
    }
    bool dfs(int u) {
        visx[u] = true;
        for (int v = 1; v <= N; v++)
			if (!visy[v]) {
				int t = lx[u] + ly[v] - g[u][v];
				if (!t) {
					visy[v] = 1;
					if (!match[v] || dfs(match[v])) {
						match[v] = u;
						return true;
					}
				}
				else sla[v] = min(sla[v], t);
            }
        return false;
    }
    Point KM() {
        memset(lx, 0, sizeof(lx));
        memset(ly, 0, sizeof(ly));
        memset(match, 0, sizeof(match));
        for (int i = 1; i <= N; i++)
            for (int j = 1; j <= N; j++)
                lx[i] = max(lx[i], g[i][j]);
        for (int i = 1; i <= N; i++) {
            memset(sla, 63, sizeof sla);
            while (1) {
                memset(visx, 0, sizeof visx);
                memset(visy, 0, sizeof visy);
                if (dfs(i)) break;
                int d = INF;
                for (int i = 1; i <= N; i++)
                    if (!visy[i]) d = min(d, sla[i]);
                for (int i = 1; i <= N; i++)
                    if (visx[i]) lx[i] -= d;
                for (int i = 1; i <= N; i++)
                    if (visy[i]) ly[i] += d;
                    else sla[i] -= d;
            }
        }
        Point res = (Point){0, 0};
        for (int i = 1; i <= N; i++)
            res.x += A[match[i]][i], res.y += B[match[i]][i];
        return res;
    }
}
void Div(Point A, Point B) {
	Gra::build(A.y - B.y, B.x - A.x);
	Point C = Gra::KM();
	ans = min(ans, C.x * C.y);
	if (cross(B - A, C - A) >= 0) return ;
	Div(A, C), Div(C, B);
}
int main () {
#ifndef ONLINE_JUDGE
    freopen("cpp.in", "r", stdin);
#endif
	int T; scanf("%d", &T);
	while (T--) {
		scanf("%d", &N);
		for (int i = 1; i <= N; i++)
			for (int j = 1; j <= N; j++)
				scanf("%d", &Gra::A[i][j]);
		for (int i = 1; i <= N; i++)
			for (int j = 1; j <= N; j++)
				scanf("%d", &Gra::B[i][j]);
		Gra::build(1, 0);
		Point A = Gra::KM();
		Gra::build(0, 1);
		Point B = Gra::KM();
		ans = min(A.x * A.y, B.x * B.y);
		Div(A, B);
		printf("%d\n", ans);
	}
    return 0;
}