题意:
If two point such as (xi,yi,zi) and (xj,yj,zj) xi≥xj yi≥yj zi≥zj, the bigger one level add 1
问每个point的level是多少。
思路：
cdq分治，先去重并统计相同的点的数量，需要注意要记录原id对应到了哪个新id
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/* *********************************************** Author :111qqz Created Time :2017年10月10日 星期二 19时53分38秒 File Name :5618.cpp ************************************************ */ #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <cmath> #include <cstdlib> #include <ctime> #define PB push_back #define fst first #define sec second #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define ms(a,x) memset(a,x,sizeof(a)) typedef long long LL; #define pi pair < int ,int > #define MP make_pair using namespace std; const double eps = 1E-8; const int dx4[4]={1,0,0,-1}; const int dy4[4]={0,-1,1,0}; const int inf = 0x3f3f3f3f; const int N=1E5+7; int n; struct point { int x,y,z; int id; int cnt,sum; void input( int _id) { scanf("%d %d %d",&x,&y,&z); id=_id; } bool operator < (const point &b)const { if (x!=b.x) return x<b.x; if (y!=b.y) return y<b.y; return z<b.z; } bool operator !=(const point &b)const { return x!=b.x||y!=b.y||z!=b.z; } }p[N]; struct BIT { int n,t[N]; void init( int _n) { n = _n; ms(t,0); } inline int lowbit(int x){ return x&(-x);} void update( int x,int delta) { for ( int i = x ; i <= n ; i+=lowbit(i)) t[i] +=delta; } int Sum( int x) { int ret = 0 ; for ( int i = x ; i >=1 ; i-=lowbit(i)) ret+=t[i]; return ret; } }bit; bool cmpcdq(point a,point b){return a.y<b.y;} bool cmpid(point a,point b){return a.id < b.id;} void cdq( int l,int r) { if (l==r) return; int mid = (l+r)>>1; cdq(l,mid); cdq(mid+1,r); sort(p+l,p+mid+1,cmpcdq); sort(p+mid+1,p+r+1,cmpcdq); int j = l; for ( int i = mid + 1 ; i <= r ; i++) { for ( ; j <= mid && p[j].y <= p[i].y ; j++) bit.update(p[j].z,p[j].cnt); p[i].sum+=bit.Sum(p[i].z); } for ( int i = l ; i < j ; i++) bit.update(p[i].z,-p[i].cnt); } int tot; int ans[N]; int id[N]; //记录原id对应到了哪个新id int main() { #ifndef ONLINE_JUDGE freopen("./in.txt","r",stdin); #endif int T; scanf("%d",&T); while (T--) { tot=0; ms(p,0); //多组数据orz scanf("%d",&n); bit.init(N-1); for ( int i = 1 ; i <= n ; i++) p[i].input(i); int cnt = 0; sort(p+1,p+n+1); for ( int i = 1 ; i <= n ; i++) { cnt++; if (p[i]!=p[i+1]) { p[i].cnt = cnt; id[p[i].id]=tot+1; p[++tot]=p[i]; p[tot].id = tot; cnt=0; } else { id[p[i].id] = tot+1; } } cdq(1,tot); //for ( int i = 1 ; i <= n ; i++) ans[id[p[i].id]] = p[i].sum ; //for ( int i = 1 ; i <= n ; i++) printf("%d\n",ans[i]); sort(p+1,p+tot+1,cmpid); for ( int i = 1 ; i <= n ; i++) printf("%d\n",p[id[i]].sum+p[id[i]].cnt-1); } #ifndef ONLINE_JUDGE fclose(stdin); #endif return 0; } |
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