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http://codeforces.com/problemset/problem/12/d
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.
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 the answer to the problem.
3 1 4 2 4 3 2 2 5 3
1
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<set>
#include<vector>
#include<map>
#include<math.h>
#include<queue>
#include<string>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define N 500005
#define ll int
ll n;
ll c[N], maxn;
inline ll lowbit(ll x){return x&(-x);}
void change(ll pos, ll val){
while(pos)c[pos]=max(c[pos],val), pos-=lowbit(pos);
}
ll maxx(ll pos){
ll ans = -1;
while(pos<=maxn)ans = max(ans,c[pos]),pos+=lowbit(pos);
return ans;
}
struct node{
ll b[3],num;
}w[N];
bool cmp0(node x, node y){return x.b[0]<y.b[0];}
bool cmp1(node x, node y){return x.b[1]>y.b[1];}
int main(){
ll i,j;
while(cin>>n) {
for(i=0;i<n;i++)scanf("%d",&w[i].b[0]);
for(i=0;i<n;i++)scanf("%d",&w[i].b[1]);
for(i=0;i<n;i++)scanf("%d",&w[i].b[2]);
sort(w, w+n, cmp0);
ll rank = 1;
w[0].num = 1;
for(i=1;i<n;i++) {
if(w[i].b[0]==w[i-1].b[0])w[i].num = rank;
else w[i].num = ++rank;
}
sort(w,w+n,cmp1);
for(i=1;i<=rank;i++)c[i]=-1;
maxn = rank;
i = 0;
ll ans = 0;
while(i<n) {
for(j = i; j < n && w[i].b[1] == w[j].b[1]; j++)
if(maxx(w[j].num+1)>w[j].b[2])
ans++;
for(j = i; j < n && w[i].b[1] == w[j].b[1]; j++)
change(w[j].num, w[j].b[2]);
i = j;
}
cout<<ans<<endl;
}
return 0;
}版权声明：本文为博主原创文章，未经博主允许不得转载。
树状数组模拟3个元素的排序 Codeforces 12D Ball
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原文地址：http://blog.csdn.net/acm_10000h/article/details/47986011
