HDU 5046 Airport
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;
}