D. Ball CF

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N ladies attend the ball in the King's palace. Every lady can be described with three values: beauty, intellect and richness. King's Master of Ceremonies knows that ladies are very special creatures. If some lady understands that there is other lady at the ball which is more beautiful, smarter and more rich, she can jump out of the window. He knows values of all ladies and wants to find out how many probable self-murderers will be on the ball. Lets denote beauty of thei-th lady by B_i, her intellect byI_i and her richness byR_i. Theni-th lady is a probable self-murderer if there is somej-th lady thatB_i < B_j, I_i < I_j, R_i < R_j. Find the number of probable self-murderers.

Input

The first line contains one integer N (1 ≤ N ≤ 500000). The second line containsN integer numbersB_i, separated by single spaces. The third and the fourth lines contain sequencesI_i andR_i in the same format. It is guaranteed that0 ≤ B_i, I_i, R_i ≤ 10⁹.

Output

Output the answer to the problem.

Sample test(s)

Input

31 4 24 3 22 5 3

Output

http://codeforces.com/problemset/problem/12/D

//这三维 一维是插入顺序 一维是插入位置 一维是插入的值#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define LL(x) (x<<1)#define RR(x) ((x<<1)|1)const int maxn=500050;struct node{    int b,i,r;}p[maxn];int num[maxn];int tot;int Bin(int x){    int l=0,r=tot,mid;    while(l<=r)    {        mid=(l+r)>>1;        if(num[mid]<x)            l=mid+1;        else if(num[mid]>x)            r=mid-1;        else            return mid;    }    return -1;}int cmp(node x,node y){    if(x.b!=y.b) return x.b<y.b;    if(x.i!=y.i) return x.i<y.i;    return x.r<y.r;}struct Seg{    int l,r;    int max;}tree[maxn*3];void build(int l,int r,int k){    tree[k].l=l;    tree[k].r=r;    tree[k].max=-1;    if(l==r) return ;    int mid=(l+r)>>1;    build(l,mid,LL(k));    build(mid+1,r,RR(k));}int Max(int a,int b){    if(a>b) return a;    return b;}void update(int l,int k,int val){    if(l==tree[k].l && l==tree[k].r)    {        tree[k].max=Max(tree[k].max,val);        return ;    }    int mid=(tree[k].l+tree[k].r)>>1;    if(l<=mid) update(l,LL(k),val);    else update(l,RR(k),val);    tree[k].max=Max(tree[LL(k)].max,tree[RR(k)].max);}int query(int l,int r,int k){    if(l>r) return -1;    if(l<=tree[k].l && tree[k].r<=r)        return tree[k].max;    int mid=(tree[k].l+tree[k].r)>>1;    if(r<=mid) return query(l,r,LL(k));    else if(l>mid) return query(l,r,RR(k));    else    return Max(query(l,r,LL(k)),query(l,r,RR(k)));}int main(){    int n;    int i,j;    while(scanf("%d",&n)!=EOF)    {        tot=0;        for(i=0;i<n;i++)            scanf("%d",&p[i].b);        for(i=0;i<n;i++)        {            scanf("%d",&p[i].i);            num[tot++]=p[i].i;        }        for(i=0;i<n;i++)            scanf("%d",&p[i].r);        sort(num,num+tot);        for(i=1,j=0;i<tot;i++)            if(num[j]!=num[i])                num[++j]=num[i];        tot=j;        for(i=0;i<n;i++)            p[i].i=Bin(p[i].i);        sort(p,p+n,cmp);        int ans=0;        build(0,tot,1);    //  for(i=0;i<n;i++)    //  {    //      printf("%d %d %d\n",p[i].b,p[i].i,p[i].r);    //  }        for(i=n-1,j=n-1;i>=0;i--)        {            while(p[j].b!=p[i].b && j>=0)            {                update(p[j].i,1,p[j].r);                j--;            }            if(p[i].r<query(p[i].i+1,tot,1))            {            //  printf("i==%d\n",i);                ans++;            }        }        printf("%d\n",ans);    }    return 0;}

（not mine）

//树状数组 其实就是二叉树的性质求 （x+1->tot  tot定值）#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 500010;int c[N];int num[N];int n;struct node{    int x,y,z;}in[N];inline int Max(int &a,int b){    if(a<b) return b;    return a;}void update(int x,int d){    for(;x<N;x+=x&-x)   c[x]=Max(c[x],d);}int sum(int x){    int ans=0;    for(;x>0;x-=x&-x)   ans=Max(ans,c[x]);    return ans;}void init(){     sort(num+1,num+n+1);     int tot=unique(num+1,num+n+1)-num-1;     for(int i=1;i<=n;i++)         in[i].x=tot-(lower_bound(num+1,num+1+tot,in[i].x)-num-1);}inline int cmp(node a,node b){    return a.y>b.y;}int main(){    int i,j,k;    scanf("%d",&n);    for(i=1;i<=n;i++) {scanf("%d",&in[i].x);    num[i]=in[i].x;}    for(i=1;i<=n;i++) scanf("%d",&in[i].y);    for(i=1;i<=n;i++) scanf("%d",&in[i].z);    init();    memset(c,0,sizeof(c));    sort(in+1,in+n+1,cmp);    int ans=0;    for(i=1;i<=n;i=j){        for(j=i;j<=n&&in[j].y==in[i].y;j++){            int z=sum(in[j].x-1);            if(z>in[j].z) ans++;        }        for(;i<=j-1;i++)            update(in[i].x,in[i].z);    }    printf("%d\n",ans);    return 0;}