CF1027D Mouse Hunt 思维

Mouse Hunt

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Medicine faculty of Berland State University has just finished their admission campaign. As usual, about 80%80% of applicants are girls and majority of them are going to live in the university dormitory for the next 44 (hopefully) years.

The dormitory consists of nn rooms and a single mouse! Girls decided to set mouse traps in some rooms to get rid of the horrible monster. Setting a trap in room number ii costs cici burles. Rooms are numbered from 11 to nn.

Mouse doesn't sit in place all the time, it constantly runs. If it is in room ii in second tt then it will run to room aiai in second t+1t+1 without visiting any other rooms inbetween (i=aii=ai means that mouse won't leave room ii). It's second 00 in the start. If the mouse is in some room with a mouse trap in it, then the mouse get caught into this trap.

That would have been so easy if the girls actually knew where the mouse at. Unfortunately, that's not the case, mouse can be in any room from 11 to nn at second 00.

What it the minimal total amount of burles girls can spend to set the traps in order to guarantee that the mouse will eventually be caught no matter the room it started from?

Input

The first line contains as single integers nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of rooms in the dormitory.

The second line contains nn integers c1,c2,…,cnc1,c2,…,cn (1≤ci≤1041≤ci≤104) — cici is the cost of setting the trap in room number ii.

The third line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n) — aiai is the room the mouse will run to the next second after being in room ii.

Output

Print a single integer — the minimal total amount of burles girls can spend to set the traps in order to guarantee that the mouse will eventually be caught no matter the room it started from.

Examples

input

Copy

5
1 2 3 2 10
1 3 4 3 3

output

Copy

3

input

Copy

4
1 10 2 10
2 4 2 2

output

Copy

10

input

Copy

7
1 1 1 1 1 1 1
2 2 2 3 6 7 6

output

Copy

2

Note

In the first example it is enough to set mouse trap in rooms 11 and 44. If mouse starts in room 11 then it gets caught immideately. If mouse starts in any other room then it eventually comes to room 44.

In the second example it is enough to set mouse trap in room 22. If mouse starts in room 22 then it gets caught immideately. If mouse starts in any other room then it runs to room 22 in second 11.

Here are the paths of the mouse from different starts from the third example:

• 1→2→2→…1→2→2→…;
• 2→2→…2→2→…;
• 3→2→2→…3→2→2→…;
• 4→3→2→2→…4→3→2→2→…;
• 5→6→7→6→…5→6→7→6→…;
• 6→7→6→…6→7→6→…;
• 7→6→7→…7→6→7→…;

So it's enough to set traps in rooms 22 and 66.

题意：一个寝室里有n个房间和一个老鼠，老鼠一开始可能在任意一个房间，老鼠会从房间i跳到a[i]房间，问在哪些房间下陷阱可以用最小的代价抓到老鼠？

分析：遍历所有房间，给从这些可能到达的房间打上标记，每次打标志在花费最小的房间布下陷阱

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e6+10;
const ll mod = 1e9+7;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll n, ans, a[maxn], c[maxn], vis[maxn];
int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    cin >> n;
    for( ll i = 1; i <= n; i ++ ) {
        cin >> c[i];
    }
    for( ll i = 1; i <= n; i ++ ) {
        cin >> a[i];
    }
    for( ll i = 1; i <= n; i ++ ) {
        ll x = i, v, mn;
        while( !vis[x] ) {  // 找到已被访问过的下一个房间
            vis[x] = i, x = a[x];
        }
        if( vis[x] != i ) {
            continue;
        }
        v = x, mn = c[x];
        while( a[x] != v ) { //在可能经过的所有房间需要花费最小的房间下陷阱
            x = a[x], mn = min( mn, c[x] );
        }
        ans += mn;
    }
    cout << ans << endl;
    return 0;
}