[Codeforces 297E]Mystic Carvings

Posted 2020-10-28 NaVi_Awson

Description

题库链接

题面链接

Solution

这里给出主席树的版本。主席树维护直线的一个端点在前 \\(i\\) 个端点中，另一个端点在区间内的个数。

Code

//It is made by Awson on 2018.3.21
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar(\'\\n\'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 2e5;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == \'-\')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(LL x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(LL x) {if (x < 0) putchar(\'-\'); print(Abs(x)); }

int n, a[N+5], b[N+5], to[N+5], len;
struct Segment_tree {
    int root[N+5], ch[N*40+5][2], key[N*40+5], pos;
    int cpynode(int o) {++pos; ch[pos][0] = ch[o][0], ch[pos][1] = ch[o][1], key[pos] = key[o]; return pos; }
    void insert(int &o, int l, int r, int loc) {
    o = cpynode(o); ++key[o];
    if (l == r) return; int mid = (l+r)>>1;
    if (loc <= mid) insert(ch[o][0], l, mid, loc);
    else insert(ch[o][1], mid+1, r, loc);
    }
    int query(int o, int l, int r, int a, int b) {
    if (!o || (a <= l && r <= b)) return key[o]; int mid = (l+r)>>1, c1 = 0, c2 = 0;
    if (a <= mid) c1 = query(ch[o][0], l, mid, a, b);
    if (b > mid) c2 = query(ch[o][1], mid+1, r, a, b);
    return c1+c2;
    }
    int query(int a, int b, int l, int r) {
    if (r < l || a > b) return 0;
    return query(root[b], 1, len, l, r)-query(root[a-1], 1, len, l, r);
    }
}T;

void work() {
    read(n);
    for (int i = 1; i <= n; i++) {
    read(a[i]), read(b[i]); to[a[i]] = b[i], to[b[i]] = a[i];
    if (a[i] > b[i]) Swap(a[i], b[i]);
    }
    len = (n<<1);
    for (int i = 1; i <= len; i++) T.root[i] = T.root[i-1], T.insert(T.root[i], 1, len, to[i]);
    LL ans = 1ll*n*(n-1)*(n-2)/2/3, t = 0;
    for (int i = 1; i <= n; i++) {
    int x = T.query(a[i]+1, b[i]-1, 1, a[i]-1)+T.query(a[i]+1, b[i]-1, b[i]+1, len);
    t += 1ll*x*(n-1-x);
    }
    ans -= t/2;
    for (int i = 1; i <= n; i++) {
    int x = T.query(a[i]+1, b[i]-1, a[i]+1, b[i]-1)/2;
    int y = (T.query(1, a[i]-1, 1, a[i]-1)+T.query(1, a[i]-1, b[i]+1, len)+T.query(b[i]+1, len, 1, a[i]-1)+T.query(b[i]+1, len, b[i]+1, len))/2;
    ans -= 1ll*x*y;
    }
    writeln(ans);
}
int main() {work(); return 0; }