| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 302777 | mhy908 | 슈퍼트리 잇기 (IOI20_supertrees) | C++14 | 272 ms | 30200 KiB |
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>
#define mp make_pair
#define eb emplace_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define svec(x) sort(all(x))
#define press(x) x.erase(unique(all(x)), x.end());
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef pair<int, LL> pil;
typedef pair<LL, int> pli;
typedef pair<LL, LL> pll;
const int INF=1e9;
const LL LLINF=1e18;
int n, arr[1010][1010], ans[1010][1010], tnum[1010];
struct UNION_FIND{
int par[1010];
UNION_FIND(){for(int i=1; i<=1000; i++)par[i]=i;}
int findpar(int a){return a==par[a]?a:par[a]=findpar(par[a]);}
void mergepar(int a, int b){par[findpar(a)]=findpar(b);}
}cyc, tree;
vector<int> rt[1010];
int construct(vector<vector<int> > p){
n=p.size();
for(int i=0; i<n; i++){
for(int j=0; j<n; j++)arr[i+1][j+1]=p[i][j];
}
for(int i=1; i<=n; i++){
for(int j=1; j<=n; j++){
if(arr[i][j]==3)return 0;
}
if(arr[i][i]!=1)return 0;
}
for(int i=1; i<=n; i++){
for(int j=1; j<=n; j++){
if(arr[i][j]==1&&i!=j)tree.mergepar(i, j);
}
}
for(int i=1; i<=n; i++){
for(int j=1; j<=n; j++){
if(tree.findpar(i)==tree.findpar(j)&&arr[i][j]!=1)return 0;
}
}
for(int i=1; i<=n; i++)tnum[i]=tree.findpar(i);
for(int i=1; i<=n; i++){
for(int j=1; j<=n; j++){
if(arr[i][j]==2)cyc.mergepar(tnum[i], tnum[j]);
}
}
for(int i=1; i<=n; i++){
for(int j=1; j<=n; j++){
if(tnum[i]==tnum[j])continue;
if(cyc.findpar(tnum[i])==cyc.findpar(tnum[j])&&arr[i][j]!=2)return 0;
}
}
for(int i=1; i<=n; i++){
rt[cyc.findpar(tnum[i])].eb(tnum[i]);
if(i==tnum[i])continue;
ans[i][tnum[i]]=1;
ans[tnum[i]][i]=1;
}
for(int i=1; i<=n; i++){
svec(rt[i]);
press(rt[i]);
if(rt[i].size()<=1)continue;
for(int j=0; j<rt[i].size()-1; j++){
ans[rt[i][j]][rt[i][j+1]]=1;
ans[rt[i][j+1]][rt[i][j]]=1;
}
ans[rt[i][0]][rt[i].back()]=1;
ans[rt[i].back()][rt[i][0]]=1;
}
vector<vector<int> > ret;
for(int i=1; i<=n; i++){
vector<int> vc;
vc.resize(n);
for(int j=1; j<=n; j++){
if(ans[i][j])vc[j-1]=1;
}
ret.push_back(vc);
}
build(ret);
return 1;
}
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