BZOJ 2648 / 2716 K-D Tree 模板题

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <ctime>
#include <cstdlib>
using namespace std;
typedef long long ll;
#define enter putchar('\n')
#define space putchar(' ')
template <class T>
char c;
bool op = 0;
while(c = getchar(), c > '9' || c < '0')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}

const int N = 1000005, D = 2, INF = 0x3f3f3f3f;
int n, m, tot, curD, root, ans;
int co[N][D], mi[N][D], ma[N][D], ls[N], rs[N];
struct point {
int co[D];
bool operator < (const point &b) const {
return co[curD] < b.co[curD];
}
} a[N];
void update(int fa, int son) {
for(int i = 0; i < D; i++) {
mi[fa][i] = min(mi[fa][i], mi[son][i]);
ma[fa][i] = max(ma[fa][i], ma[son][i]);
}
}
int build(int l, int r, int d) {
curD = d;
int mid = (l + r) >> 1;
nth_element(a + l, a + mid, a + r + 1);
for(int i = 0; i < D; i++)
co[mid][i] = mi[mid][i] = ma[mid][i] = a[mid].co[i];
if(mid > l) ls[mid] = build(l, mid - 1, (d + 1) % D), update(mid, ls[mid]);
if(mid < r) rs[mid] = build(mid + 1, r, (d + 1) % D), update(mid, rs[mid]);
return mid;
}
void insert(int k, int x, int d){
if(co[x][d] <= co[k][d]){
if(!ls[k]) ls[k] = x;
else insert(ls[k], x, (d + 1) % D);
update(k, ls[k]);
}
else{
if(!rs[k]) rs[k] = x;
else insert(rs[k], x, (d + 1) % D);
update(k, rs[k]);
}
}
int mindis(int k, const point &a) {
int ret = 0;
for(int i = 0; i < D; i++)
ret += max(mi[k][i] - a.co[i], 0) + max(a.co[i] - ma[k][i], 0);
return ret;
}
int getdis(int k, const point &a) {
int ret = 0;
for(int i = 0; i < D; i++)
ret += abs(co[k][i] - a.co[i]);
return ret;
}
void query(int k, const point &a) {
ans = min(ans, getdis(k, a));
int ld = INF, rd = INF;
if(ls[k]) ld = mindis(ls[k], a);
if(rs[k]) rd = mindis(rs[k], a);
if(ld <= rd){
if(ld <= ans) query(ls[k], a);
if(rd <= ans) query(rs[k], a);
}
else{
if(rd <= ans) query(rs[k], a);
if(ld <= ans) query(ls[k], a);
}
}
int main() {
for(int i = 1; i <= n; i++)
for(int j = 0; j < D; j++)
root = build(1, n, 0);
while(m--){
int op;
if(op == 1) {
for(int i = 0; i < D; i++)
co[tot][i] = mi[tot][i] = ma[tot][i] = a[tot].co[i];
insert(root, tot, 0);
}
else {
ans = INF;
query(root, a[tot]);
write(ans), enter;
}
}
return 0;
}


posted @ 2018-01-13 13:58  胡小兔  阅读(238)  评论(0编辑  收藏  举报