HDU 5618 Jam's problem again (cdq分治+BIT)

Posted 2020-10-12 dwtfukgv

题意：给n个点，求每一个点的满足 x y z 都小于等于它的其他点的个数。

析：三维的，第一维直接排序就好按下标来，第二维按值来，第三维用数状数组维即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int maxm = 1e6 + 5;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

struct Node{
  int x, y, z, id, num;
  bool operator < (const Node &p) const{
    return x != p.x ? x < p.x : (y != p.y ? y < p.y : z < p.z);
  }

  bool operator == (const Node &p) const{
    return x == p.x && y == p.y && z == p.z;
  }
};
Node a[maxn];
int ans[maxn];
int sum[maxn];

inline int lowbit(int x){ return -x&x; }

void add(int x, int c){
  while(x < maxn){
    sum[x] += c;
    x += lowbit(x);
  }
}

int query(int x){
  int ans = 0;
  while(x){
    ans += sum[x];
    x -= lowbit(x);
  }
  return ans;
}

inline bool cmp(const Node &lhs, const Node &rhs){
  return lhs.y < rhs.y || lhs.y == rhs.y && lhs.z < rhs.z;
}

void dfs(int l, int r){
  if(l == r){  ans[a[l].id] += a[l].num-1; return ; }
  int m = l + r >> 1;
  dfs(l, m);  dfs(m+1, r);
  sort(a+l, a+m+1, cmp);
  sort(a+m+1, a+r+1, cmp);
  int j = l;
  for(int i = m+1; i <= r; ++i){
    while(j <= m && a[j].y <= a[i].y) add(a[j].z, a[j].num), ++j;
    ans[a[i].id] += query(a[i].z);
  }
  for(int i = l; i < j; ++i)  add(a[i].z, -a[i].num);
}

int id[maxn];

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);
    for(int i = 0; i < n; ++i){
      scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].z);
      a[i].id = i;
    }

    sort(a, a + n);  ms(ans, 0);
    int cnt = 0;
    for(int i = 0; i < n; ++i, ++cnt){
      a[cnt] = a[i];  a[cnt].num = 1;
      id[a[i].id] = cnt;
      for(int j = i+1; j < n && a[i] == a[j]; i = j, id[a[j].id] = cnt, ++j)
        ++a[cnt].num;
      a[cnt].id = cnt;
    }
    dfs(0, cnt-1);
    for(int i = 0; i < n; ++i)  printf("%d\n", ans[id[i]]);
  }
  return 0;
}