Problem E: Fixed Points
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 29 Solved: 11
Description
A permutation of length n is an integer sequence such that each integer from 0 to (n - 1) appears exactly once in it. For example, sequence [0, 2, 1] is a permutation of length 3
while both [0, 2, 2] and [1, 2, 3] are not.
A fixed point of a function is a point that is mapped to itself by the function. A permutation can be regarded as a bijective function. We‘ll get a definition of a fixed point in a permutation. An integer i is a fixed point of permutation a[0],
a[1], ..., a[n-1] if and only if a[i]=i . For example, permutation [0, 2, 1] has 1 fixed point and permutation [0, 1, 2] has 3 fixed points.
You are given permutation a. You are allowed to swap two elements of the permutation at most once. Your task is to maximize the number of fixed points in the resulting permutation. Note that you are allowed to make at most one swap operation.
Input
The first line contains a single integer n (1 <= n <= 10^5). The
second line contains n integers a[0], a[1], ..., a[n - 1] ----
the given permutation.
Output
Print a single integer — the maximum possible number of fixed points in the permutation after at most one swap operation.
Sample Input
5
0 1 3 4 2
Sample Output
3
HINT
题意：找出最多交换两个数的位置的序列中a [ i ] = i 的数的个数！
思路：因为最多就交换一次，ans最多也就多2， 然而只要满足了 i = a[ a[ i ] ] 就可以多2，否则多1或不变；
具体看代码（水水）：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int fun(int a[], int b[], int n)
{
int flag=0;
for(int i=0; i < n; i++)
{
if(b[i] == 0)
{
if(i == a[a[i]]) return 2;
flag++;
}
}
if(flag)return 1;
else return 0;
}
int main()
{
int n;
int a[100010], is[100010];
while(scanf("%d", &n) != EOF)
{
int ans=0;
memset(is, 0, sizeof(is));
memset(a, 0, sizeof(a));
for(int i=0; i<n; i++)
{
scanf("%d", &a[i]);
if(a[i] == i)
{
is[i] =1;
ans++;
}
}
printf("%d\n", ans+fun(a, is, n));
}
return 0;
}