标签：map main turn 题意 input sam child version inf
https://vjudge.net/problem/CodeForces-1249B2
题目描述：
The only difference between easy and hard versions is constraints.
There are nn kids, each of them is reading a unique book. At the end of any day, the ii-th kid will give his book to the pipi-th kid (in case of i=pii=pi the kid will give his book to himself). It is guaranteed that all values of pipi are distinct integers from 11 to nn (i.e. pp is a permutation). The sequence pp doesn‘t change from day to day, it is fixed.
For example, if n=6n=6 and p=[4,6,1,3,5,2]p=[4,6,1,3,5,2] then at the end of the first day the book of the 11-st kid will belong to the 44-th kid, the 22-nd kid will belong to the 66-th kid and so on. At the end of the second day the book of the 11-st kid will belong to the 33-th kid, the 22-nd kid will belong to the 22-th kid and so on.
Your task is to determine the number of the day the book of the ii-th child is returned back to him for the first time for every ii from 11 to nn.
Consider the following example: p=[5,1,2,4,3]p=[5,1,2,4,3]. The book of the 11-st kid will be passed to the following kids:
So after the fourth day, the book of the first kid will return to its owner. The book of the fourth kid will return to him for the first time after exactly one day.
You have to answer qq independent queries.
Input
The first line of the input contains one integer qq (1≤q≤10001≤q≤1000) — the number of queries. Then qq queries follow.
The first line of the query contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of kids in the query. The second line of the query contains nn integers p1,p2,…,pnp1,p2,…,pn (1≤pi≤n1≤pi≤n, all pipi are distinct, i.e. pp is a permutation), where pipi is the kid which will get the book of the ii-th kid.
It is guaranteed that ∑n≤2⋅105∑n≤2⋅105 (sum of nn over all queries does not exceed 2⋅1052⋅105).
Output
For each query, print the answer on it: nn integers a1,a2,…,ana1,a2,…,an, where aiai is the number of the day the book of the ii-th child is returned back to him for the first time in this query.
Example
6 5 1 2 3 4 5 3 2 3 1 6 4 6 2 1 5 3 1 1 4 3 4 1 2 5 5 1 2 4 3
1 1 1 1 1 3 3 3 2 3 3 2 1 3 1 2 2 2 2 4 4 4 1 4
题意就是从一列数的第i个数开始，然后移动到第ai个数，这样不断移动下去回到第i个数需要经过多少次。显然从i->i中的所有数可以
组成一个圆环，所有圆环中的元素需要移动的次数是一样的，所以对于每一个没有搜索过的数来说，在搜索过程中出现的数字需要移动的次数
是一样的，我们把这些结果记录下来，避免重复搜索，就可以做到O(n)的复杂度了
#include<set> #include<map> #include<stack> #include<queue> #include<cmath> #include<cstdio> #include<cctype> #include<string> #include<vector> #include<climits> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define endl ‘\n‘ #define max(a, b) (a > b ? a : b) #define min(a, b) (a < b ? a : b) #define mst(a) memset(a, 0, sizeof(a)) #define IOS ios::sync_with_stdio(false) #define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n") using namespace std; typedef long long ll; typedef pair<int, int> P; const double pi = acos(-1.0); const double eps = 1e-7; const ll MOD = 1e9+7; const int INF = 0x3f3f3f3f; const int _NAN = -0x3f3f3f3f; const int NIL = -1; const int maxn = 2e5+10; int arr[maxn], ans[maxn], cnt; void solve(int x, int pos) { if (x != arr[pos]) { ++cnt; solve(x, arr[pos]); } ans[pos] = cnt; } int main(void) { int t; scanf("%d", &t); while(t--) { int n; scanf("%d", &n); for (int i = 1; i<=n; ++i) scanf("%d", &arr[i]); for (int i = 1; i<=n; ++i) if (!ans[i]) { cnt = 1; solve(i, i); } for (int i = 1; i<=n; ++i) printf(i == n ? "%d\n" : "%d ", ans[i]); mst(ans); } return 0; }
codeforces - 1249B2 Books Exchange (hard version)
标签：map main turn 题意 input sam child version inf
原文地址：https://www.cnblogs.com/shuitiangong/p/12273942.html
