CodeForces12D 树状数组降维

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

题意

给N (N<=500000)个点,每个点有x,y,z ( 0<= x,y,z <=10^9 )

对于某点(x,y,z),若存在一点(x1,y1,z1)使得x1 > x && y1 > y && z1 > z 则点(x,y,z)是特殊点。

问N个点中,有多少个特殊点。

乍一看以为是裸的三位偏序问题,直接联想到了cdq分治,但是事实上这题和三位偏序有很大的差异,三位偏序问题求的是偏序的组数,但这题问的是完全被小于的个数,cdq分治上很难维护一个点是否已经被“超越”过,也不需要这么麻烦的去维护,事实上一维将x从大到小排序,一维作为树状数组上点的位置,越大的位置在越靠前,一维就是树状数组维护的前缀最大值即可。

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d
", x)
#define Prl(x) printf("%lld
",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
const double eps = 1e-9;
const int maxn = 5e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,tmp,K,cnt;
int Hash[maxn]; 
struct Node{
    int a,b,c;
}node[maxn];
bool cmp(Node a,Node b){
    return a.a > b.a;
}
int tree[maxn];
void add(int x,int y){
    for(;x <= cnt;x += x & -x) tree[x] = max(tree[x],y);
}
int getmax(int x){
    int s = 0;
    for(;x > 0;x -= x & -x) s = max(s,tree[x]);
    return s;
}
int main()
{
    Sca(N);
    For(i,1,N) scanf("%d",&node[i].a);
    For(i,1,N) scanf("%d",&node[i].b);
    For(i,1,N) scanf("%d",&node[i].c);
    For(i,1,N) Hash[i] = node[i].c;
    sort(Hash + 1,Hash + 1 + N);
    cnt = unique(Hash + 1,Hash + 1 + N) - Hash - 1;
    For(i,1,N) node[i].c = cnt + 1 - (lower_bound(Hash + 1,Hash + 1 + cnt,node[i].c) - Hash);
    sort(node + 1,node + 1 + N,cmp);
    int ans = 0;
    For(i,1,N){
        int j = i;
        while(j <= N && node[i].a == node[j].a) j++;j--;
        For(k,i,j){
            int t = getmax(node[k].c - 1);
            if(t > node[k].b) ans++;
        }
        For(k,i,j) add(node[k].c,node[k].b);
        i = j;
    }
    Pri(ans);
    #ifdef VSCode
    system("pause");
    #endif
    return 0;
}
原文地址:https://www.cnblogs.com/Hugh-Locke/p/9614848.html