[bzoj3198][Sdoi2013]spring——容斥＋哈希表-优快云博客

题目大意：

给定一些六元组，求有多少对\((i,j)\)满足\(i,j\)中恰有\(k\)对对应相同。

思路：

考虑\(\geq k\)的对数然后简单容斥。
考虑到只有六元组，于是直接枚举子集之后把那几位提取出来，单独把那几位哈希然后计算相同的对数有多少。
但是哈希冲突很大，直接手写哈希表即可。

#include<bits/stdc++.h>

#define REP(i,a,b) for(int i=a,i##_end_=b;i<=i##_end_;++i)
#define DREP(i,a,b) for(int i=a,i##_end_=b;i>=i##_end_;--i)
#define debug(x) cout<<#x<<"="<<x<<" "
#define pii pair<int,int>
#define fi first
#define se second
#define mk make_pair
#define pb push_back
typedef long long ll;

using namespace std;

void File(){
    freopen("bzoj3198.in","r",stdin);
    freopen("bzoj3198.out","w",stdout);
}

template<typename T>void read(T &_){
    _=0; T f=1; char c=getchar();
    for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
    for(;isdigit(c);c=getchar())_=(_<<1)+(_<<3)+(c^'0');
    _*=f;
}

const ll base=97;
const ll mod=100003;
const int maxn=1e5+10;
const int maxm=10+10;
int n,m,all=(1<<6)-1;
ll a[maxn][10],f[maxm],g[maxm];

struct Hash_Table{
    int S;
    vector<pii>to[maxn];
    void reset(int S0){
        S=S0;
        REP(i,0,mod-1)to[i].clear();
    }
    bool judge(int x,int y){
        REP(i,1,6)if((1<<(i-1))&S)
            if(a[x][i]!=a[y][i])
                return false;
        return true;
    }
    ll insert(int x){
        int num=0;
        REP(i,1,6)if((1<<(i-1))&S)
            num=(num*base+a[x][i])%mod;
        REP(i,0,to[num].size()-1)
            if(judge(x,to[num][i].fi)){
                ++to[num][i].se;
                return to[num][i].se-1;
            }
        to[num].pb(mk(x,1));
        return 0;
    }
}T;

ll calc(int S){
    T.reset(S);
    ll ret=0;
    REP(i,1,n)ret+=T.insert(i);
    return ret;
}

ll C(ll x,ll y){
    ll ret=1;
    REP(i,x-y+1,x)ret*=i;
    REP(i,1,y)ret/=i;
    return ret;
}

int main(){
    File();
    read(n),read(m);
    REP(i,1,n)REP(j,1,6)read(a[i][j]);

    REP(S,0,all){
        int k=__builtin_popcount(S);
        f[k]+=calc(S);
    }

    DREP(i,6,m){
        g[i]=f[i];
        REP(j,i+1,6)g[i]-=C(j,i)*g[j];
    }

    printf("%lld\n",g[m]);

    return 0;
}

转载于:https://www.cnblogs.com/ylsoi/p/10224470.html