HDU 5046 Airport

HDU 5046 Airport

题目链接

题意：给定一些机场，要求选出K个机场，使得其他机场到其他机场的最大值最小

思路：二分+DLX重复覆盖去判断即可

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;
 }