就是先看排列p，必须满足其是一个环，才满足题意。就处理出有几个环，然后把它们合起来，答案就是多少。
然后再看序列b，自己稍微画一画就会发现，如果有偶数个1肯定是不行哒，否则，它就会再置换一圈回到它自己的位置的时候，正反面的情况和最初始相同，这样怎么转都没法在每个位置烤两面。
所以两部分的答案加起来就是最后的答案。
Pavel cooks barbecue. There are n skewers, they lay on a brazier in a row, each on one of n positions. Pavel wants each skewer to be cooked some time in every of n positions in two directions: in the one it was directed originally and in the reversed direction.
Pavel has a plan: a permutation p and a sequence b1, b2, ..., bn, consisting of zeros and ones. Each second Pavel move skewer on position i to position pi, and if bi equals 1 then he reverses it. So he hope that every skewer will visit every position in both directions.
Unfortunately, not every pair of permutation p and sequence b suits Pavel. What is the minimum total number of elements in the given permutation p and the given sequence b he needs to change so that every skewer will visit each of 2n placements? Note that after changing the permutation should remain a permutation as well.
There is no problem for Pavel, if some skewer visits some of the placements several times before he ends to cook. In other words, a permutation p and a sequence b suit him if there is an integer k (k ≥ 2n), so that after k seconds each skewer visits each of the 2nplacements.
It can be shown that some suitable pair of permutation p and sequence b exists for any n.
The first line contain the integer n (1 ≤ n ≤ 2·105) — the number of skewers.
The second line contains a sequence of integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation, according to which Pavel wants to move the skewers.
The third line contains a sequence b1, b2, ..., bn consisting of zeros and ones, according to which Pavel wants to reverse the skewers.
Print single integer — the minimum total number of elements in the given permutation p and the given sequence b he needs to change so that every skewer will visit each of 2n placements.
4
4 3 2 1
0 1 1 1
2
3
2 3 1
0 0 0
1
In the first example Pavel can change the permutation to 4, 3, 1, 2.
In the second example Pavel can change any element of b to 1.
#include<cstdio>
using namespace std;
int p[200100],n,ans;
bool b[200100],vis[200100];
int main()
{
//freopen("a.in","r",stdin);
scanf("%d",&n);
for(int i=1;i<=n;++i)
scanf("%d",&p[i]);
for(int i=1;i<=n;++i)
{
if(vis[i])
continue;
++ans;
vis[i]=1;
int U=p[i];
while(U!=i)
{
vis[U]=1;
U=p[U];
}
}
int cnt=0;
for(int i=1;i<=n;++i)
{
scanf("%d",&b[i]);
cnt+=b[i];
}
printf("%d
",(ans==1 ? 0 : ans)+(cnt%2==0));
return 0;
}