hdu4347-The Closest M Points

来源：互联网发布：典型的贪心法算法编辑：程序博客网时间：2024/05/21 18:47

http://acm.hdu.edu.cn/showproblem.php?pid=4347

找距离询问点最近的m个点，标准的k-d tree的题目，k-d tree的像一个线段树，把点分成两堆，关键在询问的时候，先按照分类的那一维先询问靠近的那一堆，然后再加自己，然后看看单独这一维的距离是否和远的那一堆已经很远了，如果很远就不去遍历远的那一堆了。、

注：dfs的时候使用全局变量一定要小心，会被更改。（在query的时候使用dim没有用局部变量之前就一直错）

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#define lson (rt << 1)#define rson (rt << 1 | 1)#define Pair pair<LD, node>using namespace std;typedef  long double LD;const int maxn = 50000 + 100;const int MAXNODE = maxn * 4 + 100;const int MAXD = 20;int K, dim, n;struct node{    int a[MAXD];    bool operator < (const node & o) const{        return a[dim] < o.a[dim];    }};LD dist(node x, node y) {    LD ans = 0;    for(int i = 0; i < K; i++){        ans += (x.a[i] - y.a[i]) * (x.a[i] - y.a[i]) * 1.0;    }    return ans;}priority_queue<Pair> q;node data[maxn], t[MAXNODE];int tag[MAXNODE];int m;void build(int dep, int rt, int l, int r) {    if(l > r) {        return;    }    tag[rt] = 1;    tag[lson] = tag[rson] = 0;    dim = dep % K;    int mid = (l + r) / 2;    nth_element(data + l, data + mid, data + r + 1);    t[rt] = data[mid];    build(dep + 1, lson, l, mid - 1);    build(dep + 1, rson , mid + 1, r);}void query(node u, int dep, int rt) {    if(tag[rt] == 0) return;    int u_dim = dep % K;    int x = lson, y = rson;    if(u.a[u_dim] > t[rt].a[u_dim]) {        swap(x, y);    }    query(u, dep + 1, x);    Pair cur;    cur.first = dist(u, t[rt]);    cur.second = t[rt];    if(q.size() < m) {        q.push(cur);    } else {        if(cur.first < q.top().first) {            q.pop();            q.push(cur);        }    }    if(q.size() < m || q.top().first > (u.a[u_dim] - t[rt].a[u_dim]) * (u.a[u_dim] - t[rt].a[u_dim])) {        query(u, dep + 1, y);    }}void out() {    printf("the closest %d points are:\n", m);    node st[MAXD];    int ans = q.size();    for(int i = 0; i < ans; i++) {        st[i] = q.top().second;        q.pop();    }    for(int i = ans - 1; i >= 0; i--) {        for(int j = 0; j < K; j++) {            printf("%d%c", st[i].a[j], (j == K - 1 ? '\n' : ' '));        }    }}int main(){    while(scanf("%d %d", &n, &K) != EOF){        for(int i = 1; i <= n; i++) {            for(int j = 0; j < K; j++) {                scanf("%d", &data[i].a[j]);            }        }        build(0, 1, 1, n);        int t;        scanf("%d", &t);        for(int i = 1; i <= t; i++) {            while(!q.empty()) q.pop();            node tmp;            for(int j = 0; j < K; j++) {                scanf("%d", &tmp.a[j]);            }            scanf("%d", &m);            query(tmp, 0, 1);            out();        }    }    return 0;}

阅读全文

0 0