HDU 5046 Airport（DLX反复覆盖）

Posted 2020-06-08

HDU 5046 Airport

题目链接

题意：给定一些机场。要求选出K个机场，使得其它机场到其它机场的最大值最小

思路：二分+DLX反复覆盖去推断就可以

代码：

#include <cstdio>
#include <cstring>

using namespace std;

const int MAXNODE = 4005;
const int MAXM = 65;
const int MAXN = 65;
const int INF = 0x3f3f3f3f;

int K;

struct DLX {

	int n, m, size;

	int U[MAXNODE], D[MAXNODE], R[MAXNODE], L[MAXNODE], row[MAXNODE], col[MAXNODE];
	int H[MAXN], S[MAXM];
	int ansd, ans[MAXN];

	void init(int n, int m) {
		this->n = n;
		this->m = m;
		ansd = INF;
		for(int i = 0; i <= m; i++) {
			S[i] = 0;
			U[i] = D[i] = i;
			L[i] = i - 1;
			R[i] = i + 1;
		}
		R[m] = 0; L[0] = m; 
		size = m;
		for(int i = 1; i <= n; i++)
			H[i] = -1;
	}

	void Link(int r, int c) {
		++S[col[++size] = c];
		row[size] = r;
		D[size] = D[c];
		U[D[c]] = size;
		U[size] = c;
		D[c] = size;
		if(H[r] < 0) H[r] = L[size] = R[size] = size;
		else {
			R[size] = R[H[r]];
			L[R[H[r]]] = size;
			L[size] = H[r];
			R[H[r]] = size;
		}
	}

	void remove(int c) {
		for(int i = D[c]; i != c; i = D[i]) {
			L[R[i]] = L[i];
			R[L[i]] = R[i];
		}
	}

	void resume(int c) {
		for(int i = U[c]; i != c; i = U[i])
			L[R[i]] = R[L[i]] = i;
	}

	bool v[MAXNODE];

	int f() {
		int ret = 0;
		for(int c = R[0]; c != 0; c = R[c]) v[c] = true;
		for(int c = R[0]; c != 0; c = R[c]) {
			if(v[c]) {
				ret++;
				v[c] = false;
				for(int i = D[c]; i != c; i = D[i])
					for(int j = R[i]; j != i; j = R[j])
						v[col[j]] = false;
			}
		}
		return ret;
	}

	bool Dance(int d) {
		if(d + f() > K) return false;
		if(R[0] == 0) {
			return d <= K;
		}
		int c = R[0];
		for(int i = R[0]; i != 0; i = R[i]) {
			if(S[i] < S[c])
				c = i;
		}
		for(int i = D[c]; i != c; i = D[i]) {
			remove(i);
			for(int j = R[i]; j != i; j = R[j]) remove(j);
			ans[d] = row[i];
			if (Dance(d + 1)) return true;
			for(int j = L[i]; j != i; j = L[j]) resume(j);
			resume(i);
		}
		return false;
	}
} gao;

typedef long long ll;

int T, n;

const int N = 65;

struct City {
	ll x, y;
	void read() {
		scanf("%I64d%I64d", &x, &y);
	}
} c[N];

ll dis(City a, City b) {
	ll dx = a.x - b.x; if (dx < 0) dx = -dx;
	ll dy = a.y - b.y; if (dy < 0) dy = -dy;
	return dx + dy;
}

int main() {
	int cas = 0;
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d", &n, &K);
		for (int i = 1; i <= n; i++) c[i].read();
		ll l = 0, r = 100000000000LL;
		while (l < r) {
			ll mid = (l + r) / 2;
			gao.init(n, n);
			for (int i = 1; i <= n; i++) {
				for (int j = 1; j <= n; j++) {
					if (dis(c[i], c[j]) <= mid)
						gao.Link(i, j);
				}
			}
			if (gao.Dance(0)) r = mid;
			else l = mid + 1;
		}
		printf("Case #%d: %I64d\n", ++cas, l);
	}
	return 0;
}