제출 #321137

#제출 시각아이디문제언어결과실행 시간메모리
321137zaneyu슈퍼트리 잇기 (IOI20_supertrees)C++14
56 / 100
502 ms23012 KiB
/*input
3
1 2 2
2 1 2
2 2 1
*/
#include "supertrees.h"
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
typedef long long int ll;
#define ld double
#define pii pair<ll,ll>
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
const ll maxn=1e3+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=(1e9+7);
const ll MOD2=1000002173;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b,ll MOD){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=(res*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
int par[maxn];
int find(int x){
    if(x==par[x]) return x;
    return par[x]=find(par[x]);
}
void merge(int a,int b){
    a=find(a),b=find(b);
    if(a==b) return;
    par[a]=b;
}
vector<vector<int>> ans;
void ed(int a,int b){
    ans[a][b]=ans[b][a]=1;
}

vector<int> comp[maxn];
int construct(vector<vector<int>> p) {
    
    int n=sz(p);
    ans.resize(n);
    REP(i,n) par[i]=i,ans[i].resize(n);
    REP(i,n){
        REP(j,n){
            if(p[i][j]==3){
                return 0;
            }
            if(p[i][j]){
                merge(i,j);
            }
            else{
                if(find(i)==find(j)){
                    return 0;
                }
            }
        }
    }
    REP(i,n){
        REP(j,n){
            if(!p[i][j] and find(i)==find(j)){
                return 0;
            }
        }
    }
    REP(i,n){
        //cout<<find(i)<<'\n';
        comp[find(i)].pb(i);
    }
    REP(i,n){
        REP(j,i){
            if(p[i][j]!=1){
                continue;
            }
            REP(k,j){
                if(p[i][j]==1 and p[j][k]==1 and p[i][k]!=1){
                    return 0;
                }
            }
        }
    }
    REP(i,n) par[i]=i;
    REP(i,n){
        REP(j,n){
            if(p[i][j]==1){
                merge(i,j);
            }
        }
    }
    REP(i,n){
        if(find(i)!=i){
            ed(i,par[i]);
        }
    }

    REP(i,n){
        vector<int> rt;
        for(int x:comp[i]){
            rt.pb(find(x));
        }
        SORT_UNIQUE(rt);
        if(sz(rt)==1) continue;
        REP(j,sz(rt)){
            ed(rt[j],rt[(j+1)%sz(rt)]);
        }
    }
    build(ans);
    return 1;
}
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