제출 #873352

#제출 시각아이디문제언어결과실행 시간메모리
873352teokakabadze슈퍼트리 잇기 (IOI20_supertrees)C++17
65 / 100
169 ms26152 KiB
#include "supertrees.h"
#include<bits/stdc++.h>
#define pb push_back
using namespace std;
 
int a[1005][1005], i, j, f[1005], a2, a3, k, l, cnt, c, pr[1005];
vector<int> s, d, e;
 
int construct(std::vector<std::vector<int>> p)
{
	int n = p.size();
	std::vector<std::vector<int>> answer;
	for(i = 0; i < n; i++)
    for(j = 0; j < n; j++)
    {
		if(p[i][j] == 3) return 0;
      	if(p[i][j] != p[j][i]) return 0;
      	if(i == j && !p[i][j]) return 0;
    }
	for(i = 0; i < n; i++)
    {
        a2 = 0;
        for(j = 0; j < n; j++)
        if(p[i][j] == 2) a2 = 1;
		if(!f[i])
        {
            e.clear(); d.clear();
            if(a2) k = 2; else k = 1;
            f[i] = 1, l = i, cnt = 0;
            if(k == 2)
            for(int b = 0; b < n; b++)
            {
                if(!f[b] && p[i][b] == 1) { f[b] = 1, a[i][b] = a[b][i] = 1; if(k == 2) e.pb(b); else d.pb(b); pr[b] = i; }
                if(f[b] == 2 && p[i][b]) return 0;
            }
            d.pb(i);
            for(j = 0; j < n; j++)
            if(!f[j] && p[l][j] == k)
            {
                f[j] = 1;
                d.pb(j);
                a[l][j] = a[j][l] = 1;
                for(int b = 0; b < n; b++)
                {
                    if(!f[b] && p[j][b] == 1) { f[b] = 1, a[j][b] = a[b][j] = 1; if(k == 2) e.pb(b); else d.pb(b); pr[b] = j; }
                    if(f[b] == 2 && p[j][b]) return 0;
                }
                l = j;
                cnt++;
            }
            else if(f[j] == 2 && p[i][j]) return 0;
            if(k == 2 && cnt < 2) return 0;
            if(k == 2) a[l][i] = a[i][l] = 1;
            for(j = 0; j < n; j++)
            if(f[j] == 1) f[j] = 2;
            for(auto x : d)
            for(auto y : e)
            if(!a[x][y] && p[x][y] != k) return 0;
            for(auto x : d)
            for(auto y : d)
            if(x != y && p[x][y] != k) return 0;
            for(auto x : e)
            for(auto y : e)
            {
                if(x != y && pr[x] != pr[y] && p[x][y] != k) return 0;
                if(x != y && pr[x] == pr[y] && p[x][y] != 1) return 0;
            }
        }
 
	}
	for(i = 0; i < n; i++)
    {
        for(j = 0; j < n; j++)
        s.push_back(a[i][j]);
        answer.push_back(s);
        s.clear();
    }
	build(answer);
	return 1;
}
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