标签：hotel int object CodeForces Resort objects path 350B ski
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:
- Objects with numbers v1, v2, ..., vk - 1 are mountains and the object with number vkis 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.
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
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