大意: n個人, 5門課, 給定每個人每門課的排名, 對於每個人輸出有多少人5門課都比他差.

明顯是個5維偏序問題, 題目有保證排名均不同, 可以用bitset優化為$O(\frac{n^2}{\omega})$

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl ‘\n‘
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<‘ ‘;hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head




const int N = 3e4+10;
int n;
bitset<N> f[N][5], tmp;
int rk[N][5], a[N][5];

int main() {
	scanf("%d", &n);
	REP(i,1,n) REP(j,0,4) {
		scanf("%d", a[i]+j);
		rk[a[i][j]][j] = i;
	}
	REP(i,1,n) REP(j,0,4) {
		f[i][j] = f[i-1][j];
		f[i][j].set(rk[i][j]);
	}
	REP(i,1,n) {
		tmp = f[a[i][0]-1][0];
		REP(j,1,4) tmp &= f[a[i][j]][j];
		printf("%d\n", int(tmp.count()));
	}
}

hihocoder 1513 小Hi的煩惱 (bitset優化)