[HAOI2017] Program number - count DP, combinatorics, inclusion and exclusion

Request from \ ((0,0,0) \) come \ ((n, m, r ) \) the number of programs. \ (n-, m, R & lt \ Leq 18 is 10 ^ {} \) . Each \ (S \) may be moved to a point \ (T \) , \ (T \) is the binary representation of the coordinates as \ (1 \) bits must contain all \ (S \) the coordinates binary representation is \ (1 \) bits. Meanwhile, \ (o \ leq 10 ^ 4 \) a checkpoint can not pass. (Original title title description surface toxic)

Solution

Consider first the program number is obtained when there is no obstacle, and apparently only on each coordinate \ (1 \) number related set \ (f [i] [j ] [k] \) represents the three coordinates \ (1 \) number followed by \ (i, j, k \ ) is the number of programs, the number of combinations pretreatment violent transfer \ (C_i ^ l \ cdot f [il] [j] [k] \ to f [i] [J] [K] \) , similar to the other two dimensions

All points in ascending order for each dimension ( \ (X \) is the first keyword, \ (Y \) is the second ......), provided \ (g [i] \) represents come impairment point \ (I \) the number of programs, consider what \ (J \) can be used as a transfer point, clearly there must be \ (x_j \ subseteq x_i, y_j \ subseteq y_i, z_j \ subseteq z_i \) , then the \ (g [i] = f [ x_i] [y_i] [z_i] - \ sum g [j] \ cdot f [i \ xor \ j] \)

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

#define int long long
const int N = 20005;
const int M = 64;
const int mod = 998244353;

namespace popcnter {
    int bt[65536];
    int popcnt (int x) {
        int t=0;
        while(x>=65536){
            t+=bt[x&65535];
            x>>=16;
        }
        return t+bt[x];
    }
    void init() {
        for(int i=0;i<65536;i++) {
            bt[i]=0;
            int x=i;
            while(x) bt[i]+=(x&1),x>>=1;
        }
    }
}

using namespace popcnter;

struct point {
    int x,y,z;
    bool operator < (const point &b) {
        if(x-b.x) return x<b.x;
        if(y-b.y) return y<b.y;
        return z<b.z;
    }
} a[N];

int p,q,r,n,f[M][M][M],g[N],c[M][M];

int cor(int x) {
    return (x%mod+mod)%mod;
}

signed main() {
    ios::sync_with_stdio(false);
    init();
    cin>>p>>q>>r;
    cin>>n;
    for(int i=1;i<=n;i++) {
        cin>>a[i].x>>a[i].y>>a[i].z;
    }
    ++n;
    a[n].x=p; a[n].y=q; a[n].z=r;
    sort(a+1,a+n+1);
    f[0][0][0]=1;
    for(int i=0;i<M;i++) {
        c[i][0]=1;
        for(int j=1;j<=i;j++) {
            c[i][j]=cor(c[i-1][j-1]+c[i-1][j]);
        }
    }
    for(int i=0;i<M;i++) {
        for(int j=0;j<M;j++) {
            for(int k=0;k<M;k++) {
                if(i+j+k) {
                    for(int l=1;l<=i;l++) f[i][j][k]=cor(f[i][j][k]+f[i-l][j][k]*c[i][l]);
                    for(int l=1;l<=j;l++) f[i][j][k]=cor(f[i][j][k]+f[i][j-l][k]*c[j][l]);
                    for(int l=1;l<=k;l++) f[i][j][k]=cor(f[i][j][k]+f[i][j][k-l]*c[k][l]);
                }
            }
        }
    }
    for(int i=1;i<=n;i++) {
        g[i]=f[popcnt(a[i].x)][popcnt(a[i].y)][popcnt(a[i].z)];
        for(int j=1;j<i;j++) {
            if((a[i].x&a[j].x)==a[j].x && (a[i].y&a[j].y)==a[j].y && (a[i].z&a[j].z)==a[j].z) {
                g[i]=cor(g[i]-g[j]*f[popcnt(a[i].x^a[j].x)]
                         [popcnt(a[i].y^a[j].y)][popcnt(a[i].z^a[j].z)]);
            }
        }
    }
    cout<<g[n];
}