The Bottom of a Graph(tarjan + 缩点)
The Bottom of a Graph
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 9139 | Accepted: 3794 |
Description
We will use the following (standard) definitions from graph theory. Let V be a nonempty and finite set, its elements being called vertices (or nodes). Let E be a subset of the Cartesian product V×V, its elements being called edges. Then G=(V,E) is called a directed graph.
Let n be a positive integer, and let p=(e1,...,en) be a sequence of length n of edges ei∈E such that ei=(vi,vi+1) for a sequence of vertices (v1,...,vn+1). Then p is called a path from vertex v1 to vertex vn+1in G and we say that vn+1 is reachable from v1, writing (v1→vn+1).
Here are some new definitions. A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from v, v is also reachable from w. The bottom of a graph is the subset of all nodes that are sinks, i.e., bottom(G)={v∈V|∀w∈V:(v→w)⇒(w→v)}. You have to calculate the bottom of certain graphs.
Let n be a positive integer, and let p=(e1,...,en) be a sequence of length n of edges ei∈E such that ei=(vi,vi+1) for a sequence of vertices (v1,...,vn+1). Then p is called a path from vertex v1 to vertex vn+1in G and we say that vn+1 is reachable from v1, writing (v1→vn+1).
Here are some new definitions. A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from v, v is also reachable from w. The bottom of a graph is the subset of all nodes that are sinks, i.e., bottom(G)={v∈V|∀w∈V:(v→w)⇒(w→v)}. You have to calculate the bottom of certain graphs.
Input
The input contains several test cases, each of which corresponds to a directed graph G. Each test case starts with an integer number v, denoting the number of vertices of G=(V,E), where the vertices will be identified by the integer numbers in the set V={1,...,v}. You may assume that 1<=v<=5000. That is followed by a non-negative integer e and, thereafter, e pairs of vertex identifiers v1,w1,...,ve,we with the meaning that (vi,wi)∈E. There are no edges other than specified by these pairs. The last test case is followed by a zero.
Output
For each test case output the bottom of the specified graph on a single line. To this end, print the numbers of all nodes that are sinks in sorted order separated by a single space character. If the bottom is empty, print an empty line.![](http://poj.org/images/2553_1.jpg)
![](http://poj.org/images/2553_1.jpg)
Sample Input
3 3 1 3 2 3 3 1 2 1 1 2 0
Sample Output
1 3 2
Source
参考代码这里:http://blog.csdn.net/ehi11/article/details/7884851
缩点:(这个概念也是看了别人的理解)先求有向图的强连通分量 , 如果几个点同属于一个强连通 , 那就给它们标上相同的记号 , 这样这几个点的集合就形成了一个缩点。
题目大意:求出度为0的强连通分量
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<stdio.h> 2 #include<queue> 3 #include<string.h> 4 using namespace std; 5 const int M = 50010 ; 6 int n , m ; 7 int stack [M] , top = 0 , index = 0; 8 bool instack [M] ; 9 int dfn[M] , low[M] ; 10 int cnt = 0 ; 11 vector <int> e[M] ; 12 int belong[M] ; 13 int out[M] ; 14 15 void init (int n) 16 { 17 top = 0 ; 18 cnt = 0 ; 19 index = 0 ; 20 memset (stack , -1 , sizeof(stack)) ; 21 memset (instack , 0 , sizeof(instack)) ; 22 memset (dfn , -1 , sizeof(dfn)) ; 23 memset (low , -1 , sizeof(low)) ; 24 for (int i = 0 ; i <= n ; i++) 25 e[i].clear () ; 26 memset (belong , -1 , sizeof(belong)) ; 27 memset (out , 0 , sizeof(out)) ; 28 } 29 30 void tarjan (int u) 31 { 32 int v ; 33 dfn[u] = low[u] = index++ ; 34 instack[u] = true ; 35 stack[++top] = u ; 36 for (int i = 0 ; i < e[u].size () ; i++) { 37 v = e[u][i] ; 38 if (dfn[v] == -1) { 39 tarjan (v) ; 40 low[u] = min (low[u] , low[v]) ; 41 } 42 else if (instack[v]) 43 low[u] = min (low[u] , dfn[v]) ; 44 } 45 if (low[u] == dfn[u]) { 46 cnt++ ; 47 do { 48 v = stack[top--] ; 49 instack[v] = false ; 50 belong[v] = cnt ; 51 } while (u != v) ; 52 } 53 } 54 55 int main () 56 { 57 //freopen ("a.txt" , "r" , stdin) ; 58 int u , v ; 59 while (~ scanf ("%d" ,&n)) { 60 if (n == 0) 61 break ; 62 init (n) ; 63 scanf ("%d" , &m) ; 64 while (m--) { 65 scanf ("%d%d" , &u , &v) ; 66 e[u].push_back (v) ; 67 } 68 for (int i = 1 ; i <= n ; i++) { 69 if (dfn[i] == -1) 70 tarjan (i) ; 71 } 72 for (int i = 1 ; i <= n ; i++) { 73 for (int j = 0 ; j < e[i].size () ; j++) { 74 if (belong [i] != belong[e[i][j]]) 75 out[belong[i]] ++; 76 } 77 } 78 int k = 0 ; 79 for (int i = 1 ; i <= n ; i++) { 80 if (out[belong[i]] == 0) { 81 if (k++) 82 printf (" ") ; 83 printf ("%d" , i) ; 84 } 85 } 86 puts ("") ; 87 } 88 return 0 ; 89 }