This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include<bits/stdc++.h>
#include "train.h"
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<ll,ll> pl;
typedef vector<ll> vl;
typedef vector<vl> vvl;
#define po pop_back
#define pb push_back
#define mk make_pair
#define mt make_tuple
#define lw lower_bound
#define up upper_bound
#define ff first
#define ss second
#define BOOST ios_base::sync_with_stdio(); cin.tie(0); cout.tie(0);
#define MOD 1000000007
#define MAX 1e18
#define MIN -1e18
#define rep(i,a,b) for(ll i=a;i<=b;i++)
#define per(i,a,b) for(ll i=b;i>=a;i--)
#define con continue
#define freopen freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
#define PI 3.14159265358979323846264338327950288419716939937510582097494459230781640628
#define read(x) scanf("%lld",&x);
#define print(x) printf("%lld ",x);
#define endl '\n';
// typedef tree<ll , null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
// template< typename T>
// using indexed_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
ll n,m,ans,mid,mn,mx,cnt,T,sum,h1,h2,e[1234567],b[1234567],c[1234567],d[1<<20],k,i,j,l,r,h,a[1234567],w,x,y,z;
bool used[1234567];
vector<int> v[1234567],vec,vv1,vv2;
string s1,s;
int sz[1234567],par[1234567];
int tr[5234567];
ll dx[4]={-1,1,0,0},dy[4]={0,0,-1,1},c1[123][123];
bool ok;
bool vis[5000];
void dfs(ll x){
if(ok==0)vis[x]=1;
ok=0;
for(auto u:v[x])if(vis[u]==0)dfs(u);
}
vector<int> who_wins(vector<int> a, vector<int> r, vector<int> U, vector<int> V){
int n=a.size() , m=U.size();
for(int i=0;i<m;i++){
v[U[i]].pb(V[i]);
}
for(int i=0;i<n;i++){
if(r[i]==1){
memset(vis, false, sizeof vis);
ok=true;
dfs(i);
if(vis[i]) c[i]=true;
}
}
vector<int>ans;
for(int i=0;i<n;i++){
ok=false;
memset(vis, false, sizeof vis);
dfs(i);
bool ok1=false;
for(int i=0;i<n;i++){
if(r[i] && vis[i] && c[i]) ok1=true;
}
if(ok1) ans.pb(1);
else ans.pb(0);
}
return ans;
}
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