Codeforces Round #605 (Div. 3) E. Nearest Opposite Parity

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define ull unsigned long long
 5 #define pb push_back
 6 const int maxn=2e5+10;
 7 const int mod=1e9+7;
 8 int n;
 9 int a[maxn];
10 int ans[maxn];
11 vector<int>E[maxn];
12 
13 
14 
15 int main()
16 {
17     ios::sync_with_stdio(false);
18     cin.tie(0);
19     cin>>n;
20     for(int i=1;i<=n;i++)
21         ans[i]=1e9;
22     for(int i=1;i<=n;i++)
23         cin>>a[i];
24     for(int i=1;i<=n;i++)
25     {
26         if(i+a[i]<=n&&a[i]%2!=a[i+a[i]]%2)
27             ans[i]=1;
28         if(i-a[i]>=1&&a[i]%2!=a[i-a[i]]%2)
29             ans[i]=1;
30     }
31 
32     for(int i=1;i<=n;i++)
33     {
34         if(i+a[i]<=n)
35         {
36             E[i+a[i]].pb(i);
37         }
38         if(i-a[i]>=1)
39         {
40             E[i-a[i]].pb(i);
41         }
42     }
43     queue<int>q;
44     for(int i=1;i<=n;i++)
45         if(ans[i]==1)
46         q.push(i);
47 
48     while(!q.empty())
49     {
50         int u=q.front();
51         q.pop();
52         for(auto &v:E[u])
53         {
54             if(ans[v]>ans[u]+1)
55             {
56                 ans[v]=ans[u]+1;
57                 q.push(v);
58             }
59         }
60     }
61     for(int i=1;i<=n;i++)
62     {
63         if(ans[i]==1e9)
64             ans[i]=-1;
65     }
66     for(int i=1;i<=n;i++)
67         cout<<ans[i]<<" ";
68 
69 
70 
71 
72 
73 
74 
75 }