cogs2479 偏序(CDQ套CDQ)

题目链接

思路

四维偏序
\(CDQ\)套\(CDQ\)，第一维默认有序。第二维用第一个CDQ变成有序的。并且对每个点标记上第一维属于左边还是右边。第二个\(CDQ\)处理第三维，注意两个\(CDQ\)不能用同一个数组，否则第二维就变成无序的了。最后一维用个树状数组统计答案。

代码

/*
* @Author: wxyww
* @Date:   2019-02-16 16:39:12
* @Last Modified time: 2019-02-17 08:18:59
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 50010;
ll read() {
    ll x=0,f=1;char c=getchar();
    while(c<'0'||c>'9') {
        if(c=='-') f=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9') {
        x=x*10+c-'0';
        c=getchar();
    }
    return x*f;
}
int n;
struct node {
    int a,b,c,d,opt;
}a[N],tmp[N],tt[N];
bool cmp(const node &x,const node &y) {
    return x.a <= y.a;
}
int tree[N];
int mx;
void update(int pos,int c) {
    while(pos <= mx) {
        tree[pos] += c;
        pos += pos & -pos;
    }
}
int query(int pos) {
    int ret = 0;
    while(pos) {
        ret += tree[pos];
        pos -= pos & -pos;
    }
    return ret;
}
vector<int>v;
int ans;
void cdq2(int l,int r) {
    if(r <= l) return;
    int mid = (l + r) >> 1;
    cdq2(l,mid);
    cdq2(mid + 1,r);
    int L = l,R = mid + 1,now = l;
    while(L <= mid && R <= r) {
        if(tmp[L].c < tmp[R].c) {
            if(tmp[L].opt == 1) update(tmp[L].d,1),v.push_back(tmp[L].d);
            tt[now++] = tmp[L++];
        }
        else {
            if(tmp[R].opt == 2) ans += query(tmp[R].d - 1);
            tt[now++] = tmp[R++];
        }
    }
    while(L <= mid) tt[now++] = tmp[L++];
    while(R <= r) {
        if(tmp[R].opt == 2) ans += query(tmp[R].d - 1);
        tt[now++] = tmp[R++];
    }
    for(int i = l;i <= r;++i) tmp[i] = tt[i];
    int k = v.size();
    for(int i = 0;i < k;++i) update(v[i],-1);
    v.clear();
}
void cdq(int l,int r) {
    if(r <= l) return;
    int mid = (l + r) >> 1;
    cdq(l,mid);
    cdq(mid + 1,r);
    int L = l,R = mid + 1,now = l;
    while(L <= mid && R <= r) { 
        if(a[L].b < a[R].b) tmp[now] = a[L++],tmp[now++].opt = 1;
        else tmp[now] = a[R++],tmp[now++].opt = 2;
    }
    while(L <= mid) tmp[now] = a[L++],tmp[now++].opt = 1;
    while(R <= r) tmp[now] = a[R++],tmp[now++].opt = 2;
    for(int i = l;i <= r;++i) a[i] = tmp[i];
    cdq2(l,r);
}
int main() {
    n = read();
    for(int i = 1;i <= n;++i) mx = max(mx,(a[i].a = i));
    for(int i = 1;i <= n;++i) mx = max(mx,(a[i].b = read()));
    for(int i = 1;i <= n;++i) mx = max(mx,(a[i].c = read()));
    for(int i = 1;i <= n;++i) mx = max(mx,(a[i].d = read()));
    cdq(1,n);
    cout<<ans;
    return 0;
}