Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
Valera's finally decided to go on holiday! He packed up and headed for a ski resort.
Valera's fancied a ski trip but he soon realized that he could get lost in this new place. Somebody gave him a useful hint: the resort has nobjects (we will consider the objects indexed in some way by integers from 1 to n), each object is either a hotel or a mountain.
Valera has also found out that the ski resort had multiple ski tracks. Specifically, for each object v, the resort has at most one object u, such that there is a ski track built from object u to object v. We also know that no hotel has got a ski track leading from the hotel to some object.
Valera is afraid of getting lost on the resort. So he wants you to come up with a path he would walk along. The path must consist of objects v1, v2, ..., vk (k ≥ 1) and meet the following conditions:
- Objects with numbers v1, v2, ..., vk - 1 are mountains and the object with number vk is the hotel.
- For any integer i(1 ≤ i < k), there is exactly one ski track leading from object vi. This track goes to object vi + 1.
- The path contains as many objects as possible (k is maximal).
Help Valera. Find such path that meets all the criteria of our hero!
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of objects.
The second line contains n space-separated integers type1, type2, ..., typen — the types of the objects. If typei equals zero, then the i-th object is the mountain. If typei equals one, then the i-th object is the hotel. It is guaranteed that at least one object is a hotel.
The third line of the input contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ n) — the description of the ski tracks. If number ai equals zero, then there is no such object v, that has a ski track built from v to i. If number ai doesn't equal zero, that means that there is a track built from object ai to object i.
Output
In the first line print k — the maximum possible path length for Valera. In the second line print k integers v1, v2, ..., vk — the path. If there are multiple solutions, you can print any of them.
Sample Input
5
0 0 0 0 1
0 1 2 3 4
5
1 2 3 4 5
5
0 0 1 0 1
0 1 2 2 4
2
4 5
4
1 0 0 0
2 3 4 2
1
1
/*
竟然看漏了一个隐藏条件，每个点的入度只能为零
*/
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
int val[];
vector<int> edge[];
int vis[];
vector<int> path[];
int n;
int a;
//每个点只可能有一个入度
int main(){
// freopen("in.txt","r",stdin);
scanf("%d",&n);
for(int i=;i<=n;i++){
scanf("%d",&val[i]);
}
for(int i=;i<=n;i++){
scanf("%d",&a);
if(a){
edge[i].push_back(a);//建图
vis[a]++;
}
}
//从1开始向前找
for(int i=;i<=n;i++){
if(val[i]){
path[i].push_back(i);
if(edge[i].size()==){
continue;
}
int pos=edge[i][];
while(true){
if(val[pos]==){//如果这一步是1的话肯定是不行的
break;
}
if(vis[pos]>){//如果这一步有两个出度也是不行的
break;
}
if(edge[pos].size()==){//如果没有下一步了
path[i].push_back(pos);
break;
}
path[i].push_back(pos);
pos=edge[pos][];
}
}
}
int maxn=-;
for(int i=;i<=n;i++){
maxn=max((int)path[i].size(),maxn);
}
for(int i=;i<=n;i++){
if(path[i].size()==maxn){
printf("%d\n",maxn);
for(int j=path[i].size()-;j>=;j--){
printf(j==path[i].size()-?"%d":" %d",path[i][j]);
}
printf("\n");
break;
}
}
return ;
}