HDU

题目：给你n个点(x,y,z)，让你输出对于每个点(xi,yi,zi)满足xj<=xi&&yj<=yi&&zj<=zi的点的数目

思路：先按照x排序，然后做分治的时候针对y排序，这样会漏掉相等的点对其左边的点的影响，做一下预处理即可

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f//0x3f3f3f3fconst int maxn=1e5+50;struct Node{    int x,y,z,id;}q[maxn],tmp[maxn];bool cmp1(Node a,Node b){    if(a.x!=b.x) return a.x<b.x;    if(a.y!=b.y) return a.y<b.y;    return a.z<b.z;}bool cmp2(Node a,Node b){    if(a.y!=b.y)        return a.y<b.y;    return a.id<b.id;}struct BIT{    int n,b[maxn];    void init(int _n){        n=_n;        mm(b,0);    }    void add(int i,int val){        for(;i<=n;i+=i&(-i))            b[i]+=val;    }    int sum(int i){        int ret=0;        for(;i>0;i-=i&(-i))            ret+=b[i];        return ret;    }}bit;int ans[maxn];int n;void cdq(int L,int R){    if(L>=R) return;    int sz=0,mid=(L+R)>>1;    for(int i=L;i<=mid;i++){        tmp[sz]=q[i];        tmp[sz++].id=0;    }    for(int i=mid+1;i<=R;i++)        tmp[sz++]=q[i];    sort(tmp,tmp+sz,cmp2);    for(int i=0;i<sz;i++){        if(tmp[i].id==0) bit.add(tmp[i].z,1);        else ans[tmp[i].id]+=bit.sum(tmp[i].z);    }    for(int i=0;i<sz;i++)        if(tmp[i].id==0)            bit.add(tmp[i].z,-1);    cdq(L,mid);    cdq(mid+1,R);}int main(){//    freopen("D:\\input.txt","r",stdin);//    freopen("D:\\output.txt","w",stdout);    int T;    scanf("%d",&T);    while(T--){        scanf("%d",&n);        mm(ans,0);        int mx=1;        for(int i=1;i<=n;i++){            scanf("%d%d%d",&q[i].x,&q[i].y,&q[i].z);            q[i].id=i;            mx=max(mx,q[i].z);        }        bit.init(mx);        sort(q+1,q+n+1,cmp1);        int cnt=0;        for(int i=n-1;i>=1;i--){            if(q[i].x==q[i+1].x&&q[i].y==q[i+1].y&&q[i].z==q[i+1].z)                cnt++;            else                cnt=0;            ans[q[i].id]+=cnt;        }        cdq(1,n);        for(int i=1;i<=n;i++)            printf("%d\n",ans[i]);    }    return 0;}