# |
Submission time |
Handle |
Problem |
Language |
Result |
Execution time |
Memory |
19159 |
2016-02-19T06:05:12 Z |
Namnamseo |
두 섬간의 연결 (kriii1_2) |
C++14 |
|
37 ms |
1864 KB |
#include <cstdio>
int par[100010];
int size[100010];
typedef long long ll;
ll c2(int x){ return x*(x+1LL)/2; }
int root(int x){ return (x==par[x])?x:(par[x]=root(par[x])); }
int main()
{
int n;
scanf("%d",&n);
ll X=0, Y=0;
for(int i=1; i<=n; ++i) par[i]=i, size[i]=1;
for(int _=1; _<n; ++_){
int a;
scanf("%d",&a);
int L=a, R=root(a+1);
X += size[L]*1LL*size[R];
Y += (c2(size[L]-1)*size[R] + c2(size[R]-1)*size[L])+size[L]*1LL*size[R];
printf("%lld %lld\n",X,Y);
par[L]=R;
size[R]+=size[L];
}
return 0;
}
# |
Verdict |
Execution time |
Memory |
Grader output |
1 |
Correct |
29 ms |
1864 KB |
Output is correct |
2 |
Correct |
37 ms |
1864 KB |
Output is correct |
3 |
Correct |
0 ms |
1864 KB |
Output is correct |
4 |
Correct |
0 ms |
1864 KB |
Output is correct |
5 |
Correct |
26 ms |
1864 KB |
Output is correct |
6 |
Correct |
0 ms |
1864 KB |
Output is correct |