Codeforces 1027D 图论
题目：
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.
题意：老鼠会在n个房间内乱窜，当老鼠在第i个房间内时，它接下来会出现在ai房间。在第i个房间安防捕鼠器的代价为ci，问最少需要花多少钱安防捕鼠夹，才能保证抓住老鼠。
方法：找环。
首先，这是一幅图，有n个节点和n条边每个结点都有且仅有一条出边，必然存在着环。
第二，老鼠无论沿何种路径走，必然会走进一个环中。
由以上两点，我们可以知道，安防捕鼠夹的最好方法，就是在每个环上找一个代价最小的地方，放置捕鼠夹。
代码如下：
#include
using namespace std;
typedef long long LL;
const int maxn = 200005;
int a[maxn];
int T;
int n;
int x;
int cnt;
int c[maxn];
int visit[maxn];
int ans;
int spos;
int Min;
//深搜找环
void dfs(int pos){
if (visit[pos] == 2)
return;
if (visit[pos] == 1){
if (pos == spos){
ans += Min;
visit[pos] = 2;
} else if (spos == -1){
spos = pos;
Min = c[pos];
dfs(a[pos]);
visit[pos] = 2;
} else {
Min = min(c[pos],Min);
dfs(a[pos]);
visit[pos] = 2;
}
return;
}
visit[pos] = 1;
dfs(a[pos]);
visit[pos] = 2;
return;
}
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 (visit[i])
continue;
Min = 200005;
spos = -1;
dfs(i);
}
cout<