标签：get raw 建图 either fan useful clu space mda
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:
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[100005]; vector<int> edge[100005]; int vis[100005]; vector<int> path[100005]; 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=1;i<=n;i++){ scanf("%d",&val[i]); } for(int i=1;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=1;i<=n;i++){ if(val[i]){ path[i].push_back(i); if(edge[i].size()==0){ continue; } int pos=edge[i][0]; while(true){ if(val[pos]==1){//如果这一步是1的话肯定是不行的 break; } if(vis[pos]>1){//如果这一步有两个出度也是不行的 break; } if(edge[pos].size()==0){//如果没有下一步了 path[i].push_back(pos); break; } path[i].push_back(pos); pos=edge[pos][0]; } } } int maxn=-1; for(int i=1;i<=n;i++){ maxn=max((int)path[i].size(),maxn); } for(int i=1;i<=n;i++){ if(path[i].size()==maxn){ printf("%d\n",maxn); for(int j=path[i].size()-1;j>=0;j--){ printf(j==path[i].size()-1?"%d":" %d",path[i][j]); } printf("\n"); break; } } return 0; }
Codeforces Round #203 (Div. 2)B Resort
标签：get raw 建图 either fan useful clu space mda
原文地址：http://www.cnblogs.com/wuwangchuxin0924/p/6602283.html
