BZOJ3053 The Closest M Points

裸的KD-tree，还是在估计要不要进入子树的时候判断一下就好了，剩下都一样

判断的方法就是看现在答案个数是否小于k，和答案是否会经过分割线。

 

  1 /**************************************************************
  2     Problem: 3053
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:2164 ms
  7     Memory:3572 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cstring>
 12 #include <algorithm>
 13 #include <queue>
 14  
 15 using namespace std;
 16 typedef long long ll;
 17  
 18 const int Cnt_kd = 1e5 + 5;
 19 const int Di = 5;
 20  
 21 struct kd_node {
 22     kd_node *s[2];
 23     int p[Di];
 24 } *kd_root, mempool[Cnt_kd], *cnt_kd, *null, *ans[25];
 25  
 26 struct data {
 27     kd_node *from;
 28     ll dis;
 29     data() {}
 30     data(kd_node *_f, ll _d) : from(_f), dis(_d) {}
 31      
 32     inline bool operator < (const data &x) const {
 33         return dis < x.dis;
 34     }
 35 };
 36  
 37 priority_queue <data> h;
 38 int n, k, K;
 39 int now[Di];
 40  
 41 inline int read() {
 42     int x = 0, sgn = 1;
 43     char ch = getchar();
 44     while (ch < '0' || '9' < ch) {
 45         if (ch == '-') sgn = -1;
 46         ch = getchar();
 47     }
 48     while ('0' <= ch && ch <= '9') {
 49         x = x * 10 + ch - '0';
 50         ch = getchar();
 51     }
 52     return sgn * x;
 53 }
 54  
 55 inline ll sqr(ll x) {
 56     return x * x;
 57 }
 58  
 59 inline ll dis(int p[Di], int q[Di]) {
 60     int i;
 61     ll res = 0;
 62     for (i = 0; i < k; ++i)
 63         res += sqr(p[i] - q[i]);
 64     return res;
 65 }
 66  
 67 #define P p -> p
 68 #define Ls p -> s[0]
 69 #define Rs p -> s[1]
 70 inline void kd_new(kd_node *&p) {
 71     p = ++cnt_kd;
 72     memcpy(P, now, sizeof(P));
 73     Ls = Rs = null;
 74 }
 75  
 76 void kd_insert(kd_node *&p, int dep) {
 77     if (p == null) {
 78         kd_new(p);
 79         return;
 80     }
 81     bool d = P[dep] < now[dep];
 82     kd_insert(p -> s[d], (dep + 1) % k);
 83 }
 84  
 85 void kd_query(kd_node *p, int dep) {
 86     if (p == null) return;
 87     h.push(data(p, dis(P, now)));
 88     while ((int) h.size() > K) h.pop();
 89     bool d = P[dep] < now[dep];
 90     kd_query(p -> s[d], (dep + 1) % k);
 91     if ((int) h.size() < K || h.top().dis > sqr(now[dep] - P[dep]))
 92         kd_query(p -> s[!d], (dep + 1) % k);
 93 }
 94  
 95 void answer() {
 96     int i, j;
 97     while (!h.empty()) h.pop();
 98     kd_query(kd_root, 0);
 99     printf("the closest %d points are:\n", K);
100     while ((int) h.size() > K) h.pop();
101     i = K;
102     while (i)
103         ans[i--] = h.top().from, h.pop();
104     for (i = 1; i <= K; ++i) {
105         for (printf("%d", ans[i] -> p[0]), j = 1; j < k; ++j)
106             printf(" %d", ans[i] -> p[j]);
107         puts("");
108     }
109 }
110 #undef P
111 #undef Ls
112 #undef Rs
113  
114 int main() {
115     int i, j, Q;
116     null = mempool;
117     null -> s[0] = null -> s[1] = null;
118     while (scanf("%d%d", &n, &k) != EOF) {
119         cnt_kd = mempool;
120         kd_root = null;
121         for (i = 1;  i <= n; ++i) {
122             for (j = 0; j < k; ++j) now[j] = read();
123             kd_insert(kd_root, 0);
124         }
125         Q = read();
126         while (Q--) {
127             for (j = 0; j < k; ++j) now[j] = read();
128             K = read();
129             answer();
130         }
131     }
132     return 0;
133 }
View Code

(p.s. 感觉kd树写的不优美！！！要改！！！)

posted on 2015-02-20 21:40  Xs酱~  阅读(156)  评论(0)  编辑  收藏  举报