HDU

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题意:在三维坐标系中有n个点(x,y,z),对于某两个点i(xi,yi,zi)和j(xj,yj,zj),如果xi>xj且yi>yj且zi>zj,则i点的level加1,求每个点的level.

分析:cdq分治+树状数组.

参考代码:

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<iostream>using namespace std;#define mem(a,b) memset(a,b,sizeof(a))const int maxn = 1e5+10;int n;struct Point{    int x,y,z,id;    Point(){}    Point( int xx, int yy, int zz, int i)    {        x = xx, y = yy, z = zz, id = i;    }}p[maxn],tmp[maxn];int maxz;int res[maxn];int bit[maxn];bool cmp( Point p, Point q){    if( p.x != q.x)        return p.x < q.x;    if( p.y != q.y)        return p.y < q.y;    return p.z < q.z;}bool cmp1( Point p, Point q){    if( p.y != q.y)        return p.y < q.y;    return p.id < q.id;}inline int lowbit( int x){    return x&(-x);}void add( int x, int val){    while( x <= maxz)    {        bit[x] += val;        x += lowbit(x);    }}int sum( int x){    int ans = 0;    while( x)    {        ans += bit[x];        x -= lowbit(x);    }    return ans;}void cdq( int l, int r){    if( l == r)        return;    int mid = (l+r)>>1;    int sub = 0;    for( int i = l; i <= mid; i++)        tmp[sub++] = Point(0,p[i].y,p[i].z,0);    for( int i = mid+1; i <= r; i++)        tmp[sub++] = Point(0,p[i].y,p[i].z,p[i].id);    sort(tmp,tmp+sub,cmp1);    for( int i = 0; i < sub; i++)    {        if( tmp[i].id == 0)            add(tmp[i].z,1);        else            res[tmp[i].id] += sum(tmp[i].z);    }    for( int i = 0; i < sub; i++)        if( tmp[i].id == 0)            add(tmp[i].z,-1);    cdq(l,mid);    cdq(mid+1,r);}int main(){    int T;    scanf("%d",&T);    while( T--)    {        maxz = 0;        scanf("%d",&n);        for( int i = 1; i <= n; i++)        {            scanf("%d%d%d",&p[i].x,&p[i].y,&p[i].z);            p[i].id = i;            maxz = max(maxz,p[i].z);        }        sort(p+1,p+1+n,cmp);        mem(res,0);        int cnt = 0;        for( int i = n; i>= 1; i--)        {            if( p[i].x == p[i+1].x && p[i].y == p[i+1].y && p[i].z == p[i+1].z)                cnt++;            else                cnt = 0;            res[p[i].id] += cnt;        }        cdq(1,n);        for( int i = 1; i <= n; i++)            printf("%d\n",res[i]);    }    return 0;}