CodeForces 350B Resort(最长路)

B - Resort
Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
Submit Status Practice CodeForces
350B

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 objectsv1, 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 thei-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 numberai 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

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

题意:求一条最长路径，路径中的节点入度和出度皆为1(忽略起始点和终点，起始点的出度必须为1)，题目已经给出限定，每一个节点的入度皆小于等于1

Objects with numbers v1, v2, ..., vk - 1 are mountains and the object with number vk is the hotel.(终点只能是旅店)

For any integer i(1 ≤ i < k), there is exactly one ski track leading from object vi. This track goes to object vi + 1.(得到的路径中每个节点的入度和出度皆为1)


The path contains as many objects as possible (k is maximal).(尽可能让这个路径上的节点数最多)

从终点到起始点进行构造，理论上效率更高(代码中时从起始点向终点构造)

思路明晰:如果有固定节点的，一定要从固定节点下手

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int MAXN = 1e5 + 5;
int n;
int A[MAXN], B[MAXN], O[MAXN], par[MAXN];
bool vis[MAXN];
void init() {
memset(O, 0, sizeof(O));
memset(vis, false, sizeof(vis));
for(int i = 0; i < MAXN; i ++) par[i] = i;
}
int main() {
while(~scanf("%d", &n)) {
init();
for(int i = 1; i <= n; i ++) {
scanf("%d", &A[i]);
}
for(int i = 1; i <= n; i ++) {
scanf("%d", &B[i]);
O[B[i]] ++;//记录出度
}
for(int i = 1; i <= n; i ++) {
if(O[B[i]] <= 1) {
par[B[i]] = i;//记录可以构成路径的节点
}
}
int dz = 0, c = 1;
for(int i = 1; i <= n; i ++) {
if(vis[i]) continue;
if(O[i] <= 1) {
int t = i,z = 1;
while(par[t] != t && !vis[i]) {//每个节点有且只会访问一次，因为入度为1
if(O[t] > 1) break;
z ++,t = par[t];
vis[t] = true;
}
//用于判断最终是否为旅店
if(A[t] == 1 && z > dz) dz = max(dz, z), c = i;
}
vis[i] = true;
}
printf("%d\n", dz);
while(par[c] != c) {
printf("%d ",c);
c = par[c];
}
printf("%d\n", c);
}
return 0;
}