Submission #359294

#	Time	Username	Problem	Language	Result	Execution time	Memory
359294		ljuba	Connecting Supertrees (IOI20_supertrees)	C++17	100 / 100	285 ms	28288 KiB

supertrees

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

typedef long long ll;
typedef long double ld;

typedef vector<int> vi;
typedef vector<ll> vll;

typedef vector<vi> vvi;
typedef vector<vll> vvll;

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

typedef vector<pii> vpii;
typedef vector<pll> vpll;

typedef vector<vpii> vvpii;
typedef vector<vpll> vvpll;

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x).size()
#define fi first
#define se second

template<class T> bool ckmin(T &a, const T &b) {return a > b ? a = b, 1 : 0;}
template<class T> bool ckmax(T &a, const T &b) {return a < b ? a = b, 1 : 0;}

void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
 
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for(auto z : x) cerr << (f++ ? "," : ""), __print(z); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if(sizeof...(v)) cerr << ", "; _print(v...);}
 
#ifdef ljuba
#define dbg(x...) cerr << "LINE(" << __LINE__ << ") -> " << "[" << #x << "] = ["; _print(x)
#else
#define dbg(x...)
#endif

const char nl = '\n';

#include "supertrees.h"

const int mxN = 1005;
bool vis[mxN];
int n;
vi cur;
vvi p;
vvi ans;

struct DSU {
    int rod[mxN], vel[mxN];
    
    DSU() {
        for(int i = 0; i < mxN; ++i) {
            rod[i] = i;
            vel[i] = 1;
        }
    }

    int find_set(int a) {
        if(a == rod[a])
            return a;
        return rod[a] = find_set(rod[rod[a]]);
    }

    bool unite(int a, int b) {
        a = find_set(a);
        b = find_set(b);

        if(a == b)
            return 0;
        if(vel[a] > vel[b])
            swap(a, b);
        vel[b] += vel[a];
        rod[a] = b;
        return 1;
    }
}dsu;

void dfs(int s) {
    cur.pb(s);
    dbg(s);
    vis[s] = 1;
    for(int i = 0; i < n; ++i) {
        if(!vis[i] && p[s][i])
            dfs(i);
    }
}

bool idi(int s) {
    cur.clear();
    dfs(s);
    for(auto z : cur) {
        for(auto z2 : cur) {
            if(!p[z][z2])
                return 0;
        }
    }
    //resenje bi trebalo da izgleda kao da dole imas lance, a iznad lanaca jos dva cvora tako da su oba povezana sa pocetkom svih lanaca

    vi lanac;

    dbg(cur);

    for(auto z : cur) {
        if(dsu.find_set(z) == z) {
            lanac.pb(z);
            vi comp;
            for(int i = 0; i < n; ++i) {
                if(dsu.find_set(i) == z) {
                    comp.pb(i);
                }
            }
            dbg(comp);
            for(int i = 0; i + 1 < sz(comp); ++i) {
                ans[comp[i]][comp[i+1]] = ans[comp[i+1]][comp[i]] = 1;
            }
            for(auto z : comp) {
                for(auto z2 : comp)
                    if(p[z][z2] != 1)
                        return 0;
            }
        }
    }
    
    dbg(lanac);

    if(sz(lanac) == 2)
        return 0;
    if(sz(lanac) == 1)
        return 1;
    for(int i = 0; i < sz(lanac); ++i) {
        ans[lanac[i]][lanac[(i+1)%sz(lanac)]] = ans[lanac[(i+1)%sz(lanac)]][lanac[i]] = 1;
    }

    return 1;
}

int construct(vvi p2) {
    p = p2;
    n = sz(p);
    ans = vvi(n, vi(n, 0));

    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            if(p[i][j] == 3)
                return 0;
        }
    }

    for(int i = 0; i < n; ++i) {
        for(int j = 0; j < n; ++j) {
            if(p[i][j] == 1)
                dsu.unite(i, j);
        }
    }

    for(int i = 0; i < n; ++i) {
        if(!vis[i]) {
            if(!idi(i))
                return 0;
        }
    }

    dbg(ans);
    build(ans);
    return 1;
}

• Subtask 1: 11 / 11
• Subtask 2: 10 / 10
• Subtask 3: 19 / 19
• Subtask 4: 35 / 35
• Subtask 5: 21 / 21
• Subtask 6: 4 / 4