D. Ball

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

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

Description

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 the i-th lady by Bi, her intellect by Ii and her richness by Ri. Then i-th lady is a probable self-murderer if there is some j-th lady that Bi < Bj, Ii < Ij, Ri < Rj. Find the number of probable self-murderers.

Input

The first line contains one integer N (1 ≤ N ≤ 500000). The second line contains N integer numbers Bi, separated by single spaces. The third and the fourth lines contain sequences Ii and Ri in the same format. It is guaranteed that 0 ≤ Bi, Ii, Ri ≤ 109.

Output

Output the answer to the problem.

Sample Input

3
1 4 2
4 3 2
2 5 3

Sample Output

1

HINT

题意

如果存在A女的三个属性都比B女高，那么B女就会自杀，然后问你有多少女的自杀了

题解：

排序，之后，然后用map搞一搞

较为详细的题解：http://blog.csdn.net/qiqijianglu/article/details/8494186

代码

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1000010
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar(‘0‘);puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar(‘0‘);puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

struct node
{
    int a,b,c;

}la[maxn];
bool cmp(node aa,node bb)
{
    if(aa.a!=bb.a)return aa.a<bb.a;
    if(aa.b!=bb.b)return aa.b>bb.b;
    return aa.c<bb.c;
}
map<int,int> M;
map<int,int>::iterator it;
int main()
{
    int n=read();
    for(int i=1;i<=n;i++)
        la[i].a=read();
    for(int i=1;i<=n;i++)
        la[i].b=read();
    for(int i=1;i<=n;i++)
        la[i].c=read();
    sort(la+1,la+1+n,cmp);
    M[-inf]=inf;
    M[inf]=-inf;
    int ans=0;
    for(int i=n;i>=1;i--)
    {
        it=M.upper_bound(la[i].b);
        if(la[i].c<it->second)
            ans++;
        else
        {
            if(M[la[i].b]<la[i].c)
            {
                M[la[i].b]=la[i].c;
                for(it=--M.lower_bound(la[i].b);it->second<=la[i].c;)
                    M.erase(it--);
            }
        }
    }
    cout<<ans<<endl;
}