CF1105E Helping Hiasat Maximum Independent Set

Portal

Title description

analysis

Largest independent set

First of all, the owner between two adjacent ones can only choose one at most, so we can convert the person’s name into a number, and then connect the edges between every two people between the ones.

Then find the largest clique of the complement graph

(At first I thought it was a bipartite graph and ran Hungary, but was later reminded that this is not a bipartite graph)

Code

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <cstring>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define _CRT_SECURE_NO_WARNINGS
#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline")
#pragma GCC option("arch=native","tune=native","no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f;
const int N = 110;
int g[N][N];
int ans,cnt[N],n,m,vis[N];
map<string,int> M;
int ppp;

bool dfs(int u,int dep){
    
    
    for(int i = u + 1;i <= n;i++){
    
    
        int flag = 1;
        if(cnt[i] + dep <= ans) return 0;
        if(g[u][i]){
    
    
            for(int j = 0;j < dep;j++)
                if(!g[i][vis[j]]){
    
    
                    flag = 0;
                    break;
                }
            if(flag){
    
    
                vis[dep] = i;
                if(dfs(i,dep + 1)) return 1;
            }
        }
    }
    if(dep > ans){
    
    
        ans = dep;
        return 1;
    }
    return 0;
}

void run(){
    
    
    ans = -1;
    for(int i = n;i;i--){
    
    
        vis[0] = i;
        dfs(i,1);
        cnt[i] = ans;
    }
}

int main(){
    
    
    scanf("%d%d",&m,&n);
    vector<int> Q;
    memset(g,1,sizeof g);
    while(m--){
    
    
        int x;
        scanf("%d",&x);
        if(x == 1) Q.clear();
        else{
    
    
            string str;
            cin >> str;
            if(!M[str]) M[str] = ++ppp;
            for(int i = 0;i < Q.size();i++)
                g[Q[i]][M[str]] = g[M[str]][Q[i]] = 0;
            Q.push_back(M[str]);
        }
    }
    run();
    printf("%d\n",ans);
    return 0;
}

/**
*　　┏┓　　　┏┓+ +
*　┏┛┻━━━┛┻┓ + +
*　┃　　　　　　　┃
*　┃　　　━　　　┃ ++ + + +
*  ████━████+
*  ◥██◤　◥██◤ +
*　┃　　　┻　　　┃
*　┃　　　　　　　┃ + +
*　┗━┓　　　┏━┛
*　　　┃　　　┃ + + + +Code is far away from 　
*　　　┃　　　┃ + bug with the animal protecting
*　　　┃　 　 ┗━━━┓ 神兽保佑,代码无bug　
*　　　┃ 　　　　　　 ┣┓
*　　  ┃ 　　　　　 　┏┛
*　    ┗┓┓┏━┳┓┏┛ + + + +
*　　　　┃┫┫　┃┫┫
*　　　　┗┻┛　┗┻┛+ + + +
*/