Topcoder srm div2 500

Problem Statement

Hero has just constructed a very specific graph. He started with n isolated vertices, labeled 0 through n-1. For each vertex i Hero then chose a vertex a[i] (other than i) and he added an edge that connected i and a[i]. This way he created a graph with n vertices and n edges. Note that if a[x]=y and a[y]=x, the vertices x and y were connected by two different edges. Hero now wants to perform the following procedure: Add a new isolated vertex number n. Choose a subset M of the original vertices. For each x in M, erase an edge between vertices x and a[x]. For each x in M, add a new edge between vertices x and n. Hero's goal is to create a final graph in which the vertices 0 through n-1 are all in the same connected component. (I.e., there must be a way to reach any of these vertices from any other of them by following one or more consecutive edges, possibly visiting vertex n along the way.) Note that Hero does not care whether vertex n is in the same component as the other vertices: both possibilities are fine. In step 2 of the above procedure Hero has 2^n possible subsets to choose from. A choice of M is good if it produces a graph with the desired property. Count how many of the 2^n possibilities are good choices. Return that count as a long long.

题意是说给出一个数组a,a[i]=j表示i到j有一条边。然后选定一个点的集合，这个集合中所有点a[x]=y的边都去掉，并把这些点都连接向点n。如果最终形成的图还是联通，那么就是好集合。问这样的集合有多少个。

和CF22E很像。。。如果每一个联通分量里面点的个数为x，尾巴上的点的个数为y，那么这块的集合的个数为2^x-2^y。把每一部分乘到一起。

代码：

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