Worfzyq likes Permutation problems.Caoshen and Mengjuju are expert at these problems . They have n cards,and all of the numbers vi on these cards are different . Because Caoshen doesn't
like disordered permutations,he wants to change the permutation into non-descending
permutation.He defines the operations:every time you can choose two digits casually,and
exchange the positions of them.Caoshen is lazy,he wants to know at least how many operations he
needs to change the permutation into non-descending one?
输入
There are multiple test cases. Each case contains a positive integer n,incicate the number of cards(n<=1000000) . Followed by n positive numbers integers v1,v2,...,vn (1≤vi≤n)
---the value of each card.
输出
Print the minmum operations in a signal line for each test case.
样例输入
51 3 2 5 4
样例输出
2
#include<iostream>
#include<stdlib.h>
#include<string.h>
#include<stdio.h>
using namespace std;
const int maxn = 1e6 + 7;
struct Node{
int v, i;
}a[maxn];
int size;
int cmp(const void *m, const void *n){
return ((Node*)m)->v - ((Node*)n)->v;
}
bool vis[maxn];
int dfs(int x){
vis[x] = true;
if (vis[a[x].i])return 0;
else return 1 + dfs(a[x].i);
}
int main(){
freopen("in.txt", "r", stdin);
while (~scanf("%d", &size)){
for (int i = 0; i < size; i++)scanf("%d",& a[i].v), a[i].i = i;
qsort(a, size, sizeof(Node), cmp);
memset(vis, 0, sizeof(vis));
long long int ans = 0;
for (int i = 0; i < size; i++){
if (!vis[i])ans += dfs(i);
}
printf("%lld
", ans);
}
return 0;
}