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  • hihocoder 1513 小Hi的烦恼 (bitset优化)

    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 '
    '
    #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
    ", int(tmp.count()));
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/uid001/p/10697871.html
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