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whiteawll 2022-08-16 14:35:09 博主文章分类：练习题目 ©著作权

文章标签 分治树状数组 #define git ios 文章分类 后端开发

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// Problem: Contest
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/21763/1021
// Memory Limit: 262144 MB
// Time Limit: 2000 ms
// 2022-04-16 15:31:39
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);
inline ll ksc(ll x,ll y,ll mod)
{
    ll ans = 0;
    while (y) {
        if (y & 1)
            ans = (ans + x) %mod;
        y >>= 1;
        x = (x + x) %mod;
    }
    return ans;
}
inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a;
        a = a * a;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
// inline ll inv (ll a) {
    // return qmi(a, mod - 2);
// }
const int N = 200005;
int a[N], b[N], c[N];
int n;
int p[N];
int sum[N];
void add (int x, int i) {
    for (;x <= n; x += x & -x)
        sum[x] += i;
}
ll query (int x) {
    ll tar = 0;
    for (;x; x -= x & -x)
        tar += sum[x];
    return tar;
}
ll get (int a[], int b[]) {
    memset(sum, 0, sizeof sum);
    int tmp[N];
    for (int i = 1; i<= n; i ++)
        tmp[a[i]] = b[i];
    ll ans = 0;
    for (int i = 1; i <= n; i ++) {
        ans += query(n) - query(tmp[i]);
        //表示求第n - tmp[i] +1的贡献，
        //下标表示第几名
        add (tmp[i], 1);
    }
    return ans;
}
void solve() {
    cin >> n;
    for (int i = 1; i<= n; i ++) cin >> a[i] >> b[i] >> c[i];
    cout << (get(a, b) + get(a, c) + get(b, c)) / 2;
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}