cf 1027D Mouse Hunt

原题

D. 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.

 

分析

给你n个点,每个点有一个权重和唯一的后继。你要从中选择一个权重和最小的点集S,使得从任何点出发,最后都会到达S中的某个点。

n个点,每个点出度为1。这样的图中一定有环,而且环与环之间不相交。那么从任意一个点出发,它最后一定会到达一个环。

只需要求出有向图中所有的环,每个环贡献一个权值最小的点即可。

 

代码
/*
AUTHOR: maxkibble
LANG: c++
PROB: cf 1027D
*/

#include bits/stdc++.h

using namespace std;

const int maxn = 2e5 + 5;

int n, c[maxn], a[maxn], v[maxn], ans;
vectorint cir;

void dfs(int x) {
    v[x] = 1;
    if (v[a[x]] == 1) cir.push_back(x);
    else if (v[a[x]] == 0) dfs(a[x]);
    v[x] = 2;
}

int main() {
    scanf("%d", n);
    for (int i = 1; i = n; i++) scanf("%d", c[i]);
    for (int i = 1; i = n; i++) scanf("%d", a[i]);
    for (int i = 1; i = n; i++) if (!v[i]) dfs(i);
    for (int i: cir) {
        int x = a[i], minv = c[i];
        while (x != i) {
            minv = min(minv, c[x]);
            x = a[x];
        }
        ans += minv;
    }
    printf("%d\n", ans);
    return 0;
}

 

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