Hihocoder #1513 : 小Hi的烦恼

#1513 : 小Hi的烦恼

https://hihocoder.com/problemset/problem/1513

分析：

　　bitset，五维数点问题。

　　记录每一科的第i名前面有那些人，最后&起来就行了。

代码；

复杂度$O(k n^2/64)$，k为维数。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cctype>
#include<set>
#include<vector>
#include<queue>
#include<map>
#include<bitset>
#define fi(s) freopen(s,"r",stdin);
#define fo(s) freopen(s,"w",stdout);
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 30001;
bitset<N> f[5][N], tmp;
int rnk[N][5], ord[N];
// rnk[i][j] 第i人的第j科的排名。 ord[i] 当前这一科中，排名为i的是谁。 
int main() {
    int n = read();
    for (int i=1; i<=n; ++i) 
        for (int j=0; j<5; ++j) rnk[i][j] = read();
    for (int j=0; j<5; ++j) { 
        for (int i=1; i<=n; ++i) ord[rnk[i][j]] = i - 1;
        for (int i=1; i<=n; ++i) {
            f[j][i] = f[j][i - 1];
            f[j][i].set(ord[i]);
        }
    }
    for (int i=1; i<=n; ++i) {
        tmp.set();
        for (int j=0; j<5; ++j) tmp &= f[j][rnk[i][j] - 1];
        printf("%d
",tmp.count());
    }
    return 0;
}

分块打表，$O(k n sqrt n / 64)$

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cctype>
#include<set>
#include<vector>
#include<queue>
#include<map>
#include<bitset>
#define fi(s) freopen(s,"r",stdin);
#define fo(s) freopen(s,"w",stdout);
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 30001;
bitset<N> f[5][175], g[5], tmp;
int rnk[N][5], ord[5][N], bel[N], pos[N];

void add(int j,int l,int r) {
    for (int i=l; i<=r; ++i) g[j].set(ord[j][i]);
}

int main() { 
    int n = read();
    for (int i=1; i<=n; ++i) 
        for (int j=0; j<5; ++j) 
            rnk[i][j] = read(), ord[j][rnk[i][j]] = i - 1;
    
    int B = sqrt(n);
    for (int j=0; j<5; ++j) { 
        int p = 1;
        for (int i=1; i<=n; i+=B,++p) 
            for (int k=1; k<=i; ++k) f[j][p].set(ord[j][k]);
    }
    for (int x,i=1; i<=n; ++i) {
        if ((i - 1) % B == 0) x = (i - 1) / B + 1, pos[x] = i;
        bel[i] = x;
    }
    for (int i=1; i<=n; ++i) {
        tmp.set();
        for (int j=0; j<5; ++j) {
            int t = rnk[i][j] - 1;
            g[j] = f[j][bel[t]];
            if (pos[bel[t]] < t) add(j, pos[bel[t]] + 1, t);
            tmp &= g[j];
        }
        printf("%d
",tmp.count());
    }
    return 0;
}