이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "supertrees.h"
#include <bits/stdc++.h>
using namespace std;
int construct(vector<vector<int>> p) {
int n = p.size();
vector<vector<int>> answer;
answer.resize(n);
for(int i = 0; i < n; i++) answer[i].resize(n);
vector <int> comp;
// last subtask, p[i][j] == 3
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
if(p[i][j] > 2) return 0;
vector <bool> isroot;
int countcomp(0);
comp.resize(n);
isroot.resize(n);
for(int i = 0; i < n; i++) { // we consider i to be the root of the connected component
if(comp[i] != 0) continue; // it can't be a root
comp[i] = ++countcomp;
isroot[i] = true;
for(int j = 0; j < n; j++) {
if(i == j) continue;
if(p[i][j] == 1) { // if there is only 1 edge from i to j
comp[j] = countcomp;
answer[i][j] = answer[j][i] = 1;
}
}
}
// now we connect the roots between them to form a cycle
// we connect the i-th root with the i+1-th root
for(int i = 0; i < n; i++) {
if(isroot[i] == true) {
for(int j = i + 1; j < n; j++) {
if(p[i][j] == 2 && isroot[j]) {
answer[i][j] = answer[j][i] = 1;
j = n;
}
}
}
}
// we connect the last with the first root
for(int i = 0; i < n; i++) {
if(isroot[i] == true) {
for(int j = n - 1; j > i; j--) {
if(p[i][j] == 2 && isroot[j]) {
answer[i][j] = answer[j][i] = 1;
j = 0;
}
}
i = n;
}
}
// now we need to verify if our build is fine
// 1. We can't have a cycle of length 2
if(countcomp == 2) {
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
if(p[i][j] == 2)
return 0;
}
// 2. If p[a][b] = 1 and p[b][c] = 1, then p[a][c] = 1
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++)
if(comp[i] != comp[j] && p[i][j] == 1)
return 0;
}
build(answer);
return 1;
}
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