CodeForces - 350B(反向建图，)

B - Resort

 CodeForces - 350B 

B. Resort

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

1. Objects with numbers v1, v2, ..., vk - 1 are mountains and the object with number vk is the hotel.
2. For any integer i (1 ≤ i < k), there is exactly one ski track leading from object vi. This track goes to object vi + 1.
3. 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 aiequals 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.

Examples

input

Copy

5
0 0 0 0 1
0 1 2 3 4

output

Copy

5
1 2 3 4 5

input

Copy

5
0 0 1 0 1
0 1 2 2 4

output

Copy

2
4 5

input

Copy

4
1 0 0 0
2 3 4 2

output

Copy

1
1



　　　　　　读题就读了好长好长时间.
　　　　　题意呢，就是：有n个地方，给出n大小的序列a，0代表ai是山地，1代表ai是旅馆，再给出个n大小的序列v，第i的序列表示ai可以冲vi到达，问怎么走才能使路途最长(终点要是旅馆，而且路中途没有分叉)
　　　　
　　　　　　拿样例二来说
　　　　　　0 1 2 2 4　　　　就是2连1，3连2,4-2,5-4,画图就是这个样子：
　　　　　　交叉点不能算，所以2忽略，根据0 0  1 0 1,　　　　3，5都是旅馆，均为终点，有3和4，5两条，第二条最长所以选第二个。
　　　　　　因为题意终点是旅馆，但我们可以反向建图，从旅馆出发，走下去直到不能再走，比如遇到分岔点或者尽头，这样问题就很清晰了。由于涉及到连接，可以用并查集来存和查找，也是个dfs的过程。
　　　　

#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
using namespace std;
const int maxn=1e5+10;
int a[maxn],b[maxn],fa[maxn];
int ac[maxn];
int dfs(int i)
{
    int ans=0;　　　　　　//在以i往下走的过程中，记录步数
    while(fa[i]!=-1)
    {
        i=fa[i];
        ans++;
    }
    return ans;
}
int main()
{
    int n;
    while(cin>>n)
    {
        memset(fa,-1,sizeof(fa));　　//边界
        map<int,int>mm;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];　　　　//存入节点类型，旅馆或山头
        }
        for(int i=1;i<=n;i++)
        {
            cin>>b[i];　　　　//存入b【i】，同时计数
            mm[b[i]]++;
        }
        for(int i=1;i<=n;i++)
        {
            if(mm[i]<=1)    //如果不是分岔点，那么可以表明,i的根节点是b[i]
                fa[i]=b[i];　　//如果,mm[i]>1，则此为分岔点，fa=-1，也为边界
        }
        int maxx=-1;　　　　//记录最长路径
        int k;　　　　　　　　//记录最长路径的起始点（旅馆）
        for(int i=1;i<=n;i++)
        {
            if(a[i]==1)
            {
                int mid=dfs(i);　　//如果是旅馆，以此i跑dfs
                if(mid>maxx)
                {
                    maxx=mid;　　//记录最大步数，并记录i值
                    k=i;
                }
            }
        }
        cout<<maxx<<endl;
        int tot=1;
        while(1)
        {　　　　　　　　　　//开新数组记录过程，以k为起点，不断跑下去，记录。
            ac[tot++]=k;
            k=fa[k];　　
            if(fa[k]==-1)　　//跑到边界就break;
                break;
        }
        for(int i=tot-1;i>=1;i--)
            cout<<ac[i]<<" "; 
            cout<<endl;
    }
}

好题！妥妥的好题！！！