大意: n个人, 5门课, 给定每个人每门课的排名, 对于每个人输出有多少人5门课都比他差.
明显是个5维偏序问题, 题目有保证排名均不同, 可以用bitset优化为$O(\frac{n^2}{\omega})$
#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())); } }
原文地址:https://www.cnblogs.com/uid001/p/10697871.html
时间: 2024-10-04 12:44:29