Pavel and barbecue

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 2n placements.
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
Note
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.

题目大意：

一共有N个烤炉，我们现在有N个串串，一开始正面朝上放在N个位子上。一秒之后，在位子i的串串会移动到pi位子上，并且如果bi==1，那么我们还要将串串翻个面。
现在要求每个串串都要在每个位子的反正面都被烤过才能吃，然而现在的机制不一定能够完成任务，让你修改最少的操作，使得满足这个操作。
思路:
要使串串在每个位置都考过,就要求这些烤炉之间形成一个回路,只有这样才能使串串在每个烤炉都烤过.并且因为在一个烤炉可能只烤过一面,那么就要在循环一次重新来到这个烤炉的时候,会烤到另一面,这就要求,在一次循环中串串翻转的次数为奇数.
找到在这些烤炉中有多少个回路,把他们变为一个回路,把n个回路变成一个回路需要n次操作,除了只有一个回路时,就不用操作.在判断翻转的次数是否为偶数,如果是偶数就要把其中一个1改为0,或0改为1.

#include <stdio.h>

int a[200005],b[200005],flag[200005]={0};

int main()
{
    int x,y,z,t,n,m;
    scanf("%d",&n);
    for(x=1;x<=n;x++)
    {
        scanf("%d",&a[x]);  
    }
    z = 0;
    for(x=1;x<=n;x++)
    {
        scanf("%d",&b[x]);
        if(b[x])
        z++;
    }
    t = 0;
    for(x=1;x<=n;x++)
    {
        if(!flag[x])
        {
            t++;
            flag[x] = 1; 
            y = a[x];
            while(y!=x)
            {
                flag[y] = 1;
                y = a[y];
            }       
        }
    }
    t = t==1?0:t;
    m = t + (z+1)%2;
    printf("%d\n",m);
    return 0;
}