CodeForces - 350B Resort （DFS）

Resort

 

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 n objects (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 vkis 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 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.

Examples

Input

5
0 0 0 0 1
0 1 2 3 4

Output

5
1 2 3 4 5

Input

5
0 0 1 0 1
0 1 2 2 4

Output

2
4 5

Input

4
1 0 0 0
2 3 4 2

Output

1
1

题目链接：http://codeforces.com/problemset/problem/350/B

题目大意：有n个地点，0是一座山，1是一个酒店，下一行表示可以从a[i]点到 i 点，问在没有岔路口的情况下到达一个酒店的最长路径（不能经过酒店）

思路：反向建图，从酒店开始往外走，记录最长的路径，如果一个节点可以从多个点到达就说明这个点是岔路口不再往下搜，如果这个点是酒店也停止搜索，没有往下走的路时也停止

代码：

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
#define ll long long
const int N=100005;
int a[N],n,ans,path[N],p,cnt[N];
vector<int>e[N];
void dfs(int u,int k)
{
    if(e[u].size()==0) //如果没有往下走的路了
    {
        if(k+1>ans) //如果比之前的路径长就更新路径
        {
            ans=k+1;
            p=u;
        }
        return ;
    }
    for(int i=0; i<e[u].size(); i++)
    {
        int v=e[u][i];
        if(cnt[v]>1||a[v]==1) //如果是岔路口或者酒店停止搜索
        {
            if(k+1>ans) 
            {
                ans=k+1;
                p=u;
            }
            continue;
        }
        path[v]=u;
        dfs(v,k+1); //继续往下搜索
    }
}
int main()
{
    scanf("%d",&n);
    ans=0;
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);
    int x;
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&x);
        e[i].push_back(x); //反向建图
        cnt[x]++;  //记录是否是岔路口
    }
    for(int i=1; i<=n; i++)
        if(a[i]) //从酒店开始搜索
            dfs(i,0);
    if(p==0) //如果路径的最后一点是0要减去这个点
    printf("%d\n",ans-1);
    else printf("%d\n",ans);
    while(1)  //输出路径
    {
        if(p!=0)
            printf("%d ",p);
        if(a[p]) break; //到酒店为止
        p=path[p];
    }
    printf("\n");
    return 0;
}

 

标签：hotel,int,object,CodeForces,Resort,objects,path,350B,ski
来源： https://blog.csdn.net/chimchim04/article/details/98303835