이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
//#include<bits/extc++.h>
//__gnu_pbds
/*
is I know the division, checking will be if there a segment that intersegt and not entirly contain.
i think this is a dp problem..
every segment will have some segment that it cannot be together with.. , count the number of partition
it can be turn into count the number of way to 2-color a graph
the answer seems to be 2^x ??? why
*/
struct DSU{
vector<int> P;
int n;
void init(int _n){
n = _n;
P.resize(_n+1);
for(int i=1;i<=n;i++) P[i]=i;
}
int query(int a){
if(P[a]==a) return a;
P[a] = query(P[a]);
return P[a];
}
void join(int a,int b){
a = query(a);
b = query(b);
P[a]=b;
return;
}
};
const int MOD = 1e9+7;
const int N = 2005;
vector<int> side[N];
int vis[N];
bool dfs(int cur,int c){
vis[cur]=3-c;
bool k = 1;
for(auto i : side[cur]){
if(!vis[i]) k&=dfs(i,vis[cur]);
else if(vis[i]==vis[cur]){
return 0;
}
}
return k;
}
int main(){
ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
int n;cin>>n;
vector<pair<int,int> > seg(n+1);
for(int i=1;i<=n;i++) cin>>seg[i].first>>seg[i].second;
sort(seg.begin()+1,seg.end(),[](const pair<int,int> &a,const pair<int,int> &b){return a.second<b.second;});
DSU d;
d.init(n);
for(int i=1;i<=n;i++){
for(int j=i-1;j>=1;j--) {
if(seg[i].first>seg[j].second) continue;
if(seg[j].first>seg[i].first) continue;
side[i].push_back(j);
side[j].push_back(i);
}
}
ll ans = 1;
for(int i=1;i<=n;i++){
if(!vis[i]) {
if(!dfs(i,2)){
cout<<0<<"\n";
return 0;
}
ans = (2*ans)%MOD;
}
}
cout<<ans<<"\n";
return 0;
}
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