[Codeforces 1105E] Helping Hiasat | State compression, maximum independent set, half

Main idea:

You have registered a new account on a social networking site. This account has $n(n\le 10^5$
) Records. Either you changed theID IDonce$ID$ , or an$ID$为$s(|s|\le 40)$ friends visited your space.

You have $m(m\le 40)$ friends. Every friend will visit your space at least once. If this friend visits your space every time, your$The\; ID$ and its$The\; ID$ , like, he would happy. I beg you to make many people happy at most.

Question idea:

First consider the final state of the selection,
then it must be any two adjacent 1 1All visitor records between$1$ are mutually exclusive.

In other words, there can only be one happy visitor in these visitor records

So the last selected visitor set can be explained as follows: it is impossible for two or more 相邻1操作间visitors to exist in this set

Consider converting to graph theory to build edges on mutually exclusive relationships, so this is obviously the largest independent set. Consider a bipartite graph? Of course not, because there is no guarantee that this is a bipartite graph.

See $m$ range$m\le 40$ , so you can consider folding in half$m$

Perform state pressure dp, divide the set into two halves, consider the independent set situation of the first half set, the independent set situation of the second half set, and finally consider the independent set situation when the two sets are merged. Assume $s$ and$t$ merge, then there must be$s$ mutually exclusive point set in$t$ does not exist, so it can be compared with the first half ofssThe second half of the combination of$s$ sets must satisfy$MAXS\oplus b\left[s\right]$ subset. So only need to press$dp$ preprocessssThe maximum value of all subsets of$s\; is$ just fine

For $(m/2)\ast 2^{m/2}$ to deal with mutually exclusive point sets and state pressure dp

So the total complexity: $O\left(\right(m/2)\ast 2^{m/2})$

Code：

/*** keep hungry and calm CoolGuang!  ***/
#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline")
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#include<stdio.h>
#include<queue>
#include<algorithm>
#include<string.h>
#include<iostream>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const ll INF= 1e17+7;
const ll maxn = 2e6+700;
const ll mod= 1e9+7;
const ll up = 1e13;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a){
    
    char c=getchar();T x=0,f=1;while(!isdigit(c)){
    
    if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){
    
    x=(x<<1)+(x<<3)+c-'0';c=getchar();}a=f*x;}
ll n,m,p;
vector<int>v;
map<string,int>mp;
int t[55][55];
int vis[55];
int b[maxn],dp[maxn];
int isd[maxn],c[maxn],s[maxn];
int S[maxn],T[maxn];
int work(){
    
    
	int ans = 0;
	int half = m/2;
	int MAX  = 1<<half;
	for(int i=1;i<=half;i++){
    
    
		for(int k=half+1;k<=m;k++){
    
    
			if(t[i][k]) b[i] |= 1<<(k-half-1);
		}
	}
	for(int i=1;i<=half;i++)
		for(int k=1;k<=half;k++)
			if(t[i][k]) S[i] |= 1<<(k-1);
	for(int i=half+1;i<=m;i++)
		for(int k=half+1;k<=m;k++)
			if(t[i][k]) T[i] |= 1<<(k-half-1);

	for(int i=0;i<MAX;i++){
    
    
		int flag = 1;
		for(int k=1;k<=half;k++){
    
    
			if(i>>(k-1)&1){
    
    
				c[i]++;
				s[i] |= b[k];
			}
			if((i>>(k-1)&1)&&(S[k]&i)) flag = 0;
		}
		isd[i] = flag;
		if(isd[i]) ans = max(ans,c[i]);
	}


	ll thalf = m - half;
	MAX = 1<<thalf;
	for(int i=0;i<MAX;i++){
    
    
		int flag = 1,cot = 0;
		for(int k=half+1;k<=m;k++){
    
    
			if(i>>(k-half-1)&1) cot++;
			if((i>>(k-half-1)&1) && (T[k]&i)) flag = 0;
		}
		if(flag) dp[i] = cot,ans = max(ans,cot);
	}

	for(int i=0;i<MAX;i++){
    
    
		for(int k=1;k<=thalf;k++)
			if(i>>(k-1)&1) dp[i] = max(dp[i],dp[i^(1<<(k-1))]);
	}

	MAX = 1<<half;
	int MAXS = (1<<thalf)-1;
	for(int i=0;i<MAX;i++){
    
    
		if(isd[i]) ans = max(ans,c[i]+dp[MAXS^s[i]]);
	}
	return ans;
}
int main(){
    
    
	read(n);read(m);
	int cnt = 0;
	for(int i=1;i<=n;i++){
    
    
		int op;read(op);
		if(op == 1)
			for(int k=1;k<=m;k++) vis[k] = 0;
		else{
    
    
			string s;cin>>s;
			if(!mp[s]) mp[s] = ++cnt;
			int aim = mp[s];
			for(int k=1;k<=m;k++)
				if(vis[k] && k!=aim) t[aim][k] = t[k][aim] = 1;
			vis[aim] = 1;
		}
	}
	di(work());
	return 0;
}
/***
6 3
1
2 aa
2 bb
1
2 aa
2 cc
***/