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C. Pavel and barbecue

谷梁子昂
2023-12-01

C. Pavel and barbecue
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

Input

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.

Output

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.

Examples
input
4
4 3 2 1
0 1 1 1
output
2
input
3
2 3 1
0 0 0
output
1

...一开始想的不够细,题意理解的不够好。我以为反转是移动方向变反呢,怎么也搞不出来,结果人家是反个面,哭晕在厕所啊QAQ

就是烤肉,某人希望每个烤肉都能够到达所有的点并且是正反两面都可以到达。

一开始烤肉都是正或反。

简单点想就是首先你要到达n个点,那么你这n个点应该是一个连通块。

所以先根据p[i]求出连通块来,如果连通块=1没那么不需要操作。如果连通块的个数大于等于1,那么就需要操作数=连通块数。

然后,当你回到当前点的时候,你应该反一个面,这样你才能正反两面都到过n个点的。

所以统计一下1的个数,如果是偶数那么需要改变一次。

#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
const int MAXN=2e5+7;
const int inf =1e9;
int n;
int p[MAXN];
bool vis[MAXN];
void dfs(int i)
{
    vis[i]=1;
    if(!vis[p[i]])dfs(p[i]);
}
int main()
{
    int i;
    scanf("%d",&n);
    for(i=1; i<=n; ++i)
    {
        scanf("%d",&p[i]);
    }
    int ans=0;
    for(i=1;i<=n;++i)
    {
        if(!vis[i])
        {
            dfs(i);
            ans++;
        }
    }
    if(ans==1)ans--;
    int sum=0;
    int x;
    for(i=0;i<n;++i)
    {
        scanf("%d",&x);
        if(x)sum++;
    }
    if(sum%2==0)ans++;
    printf("%d\n",ans);
    return 0;
}




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