Submission #1235093

#	Time	Username	Problem	Language	Result	Execution time	Memory
1235093		Hanksburger	Permutation Game (APIO25_permgame)	C++20	6 / 100	2094 ms	328 KiB

permgame

#include "permgame.h"
#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > perm;
vector<int> adj[405], vec;
int ans;
bool cmp(vector<int> &vec1, vector<int> &vec2)
{
    if (vec1.size()==4)
        return 1;
    if (vec2.size()==4)
        return 0;
    return vec1.size()<vec2.size(); // beware of inequality direction
}
void computePerm(vector<int> P)
{
    int visited[405]={};
    ans=0;
    perm.clear();
    for (int i=0; i<P.size(); i++)
    {
        if (visited[i])
            continue;
        vector<int> tmp;
        int cur=i;
        while (!visited[cur])
        {
            tmp.push_back(cur);
            visited[cur]=1;
            cur=P[cur];
        }
        if (tmp.size()==1)
            ans++;
        else
            perm.push_back(tmp);
    }
    sort(perm.begin(), perm.end(), cmp); // DONT FORGET THIS NEW ADDITION
}
int Alice(int M, int E, vector<int> U, vector<int> V, int N, vector<int> P)
{
    for (int i=0; i<E; i++)
    {
        adj[U[i]].push_back(V[i]);
        adj[V[i]].push_back(U[i]);
    }
    vec.push_back(0);
    for (int i=1; i<M; i++)
    {
        if (i==1)
            vec.push_back(adj[vec[i-1]][0]);
        else
        {
            int pre=vec[i-2];
            if (adj[vec[i-1]][0]==pre)
                vec.push_back(adj[vec[i-1]][1]);
            else
                vec.push_back(adj[vec[i-1]][0]);
        }
    }
    computePerm(P);
    for (int i=0; i<M; i++)
        if (adj[i].size()>=3)
            return ans;
    int optimal;
    if (ans>=N-3)
        return ans;
    if (ans==N-4)
        optimal=N-3;
    else
        optimal=N-5;
    while (1)
    {
        int updated=0;
        for (int i=0; i<perm.size(); i++)
        {
            if (perm[i].size()==4)
            {
                //cout << "here1\n";
                updated=1;
                vector<int> tmp(4);
                tmp[vec[0]]=perm[i][0];
                tmp[vec[1]]=perm[i][1];
                tmp[vec[2]]=perm[i][2];
                tmp[vec[3]]=perm[i][3];
                int edge=Bob(tmp);
                swap(P[tmp[U[edge]]], P[tmp[V[edge]]]);
                computePerm(P);
                break;
            }
        }
        if (updated)
            continue;
        for (int i=0; i<perm.size(); i++)
        {
            if (perm[i].size()>4)
            {
                //cout << "here2\n";
                updated=1;
                vector<int> tmp(4);
                tmp[vec[0]]=perm[i][0];
                tmp[vec[1]]=perm[i][1];
                tmp[vec[2]]=perm[i][2];
                tmp[vec[3]]=perm[i][4];
                int edge=Bob(tmp);
                swap(P[tmp[U[edge]]], P[tmp[V[edge]]]);
                computePerm(P);
                break;
            }
        }
        if (updated)
            continue;
        if (perm.size()>=2 && perm[0].size()==2 && perm[1].size()==2)
        {
            //cout << "here3\n";
            updated=1;
            vector<int> tmp(4);
            tmp[vec[0]]=perm[0][0];
            tmp[vec[1]]=perm[0][1];
            tmp[vec[2]]=perm[1][0];
            tmp[vec[3]]=perm[1][1];
            int edge=Bob(tmp);
            swap(P[tmp[U[edge]]], P[tmp[V[edge]]]);
            computePerm(P);
            continue;
        }
        if (perm.size()>=2 && perm[perm.size()-2].size()==3 && perm[perm.size()-1].size()==3)
        {
            //cout << "here4\n";
            updated=1;
            vector<int> tmp(4);
            tmp[vec[0]]=perm[perm.size()-2][0];
            tmp[vec[1]]=perm[perm.size()-2][1];
            tmp[vec[2]]=perm[perm.size()-2][2];
            tmp[vec[3]]=perm[perm.size()-1][0];
            int edge=Bob(tmp);
            swap(P[tmp[U[edge]]], P[tmp[V[edge]]]);
            computePerm(P);
            continue;
        }
        if (!updated)
            break;
    }
    return optimal;
}

• Subtask 0: 0 / 0
• Subtask 1: 0 / 6
• Subtask 2: 6 / 6
• Subtask 3: 0 / 10
• Subtask 4: 0 / 24
• Subtask 5: 0 / 24
• Subtask 6: 0 / 30