POJ 2553：The Bottom of a Graph【强连通】
The Bottom of a Graph
Time Limit : 6000/3000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 14 Accepted Submission(s) : 2
Problem 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+1 in 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+1 in 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.
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Sample Input
3 3 1 3 2 3 3 1 2 1 1 2 0
Sample Output
1 3 2题目大意 若节点V所能到达的点{w}，都能反过来到达v，那我们称v是sink。 强连通+缩点 就是求极大连通分量，最后统计出度为0的点，排序后输出初度为0的分量包含的每一个点。 AC——code：#include<cstdio> #include<algorithm> #include<cstring> #include<vector> #include<stack> #define min(a,b) a>b?b:a #define MAXN 5000 using namespace std; int dfn[MAXN],low[MAXN],sccno[MAXN],instack[MAXN],head[MAXN]; int num,scc_cnt,dfs_clock; vector<int>scc[MAXN]; stack<int>s; struct Eage { int from,to,next; }eage[MAXN*MAXN]; void add(int a,int b) { Eage e={a,b,head[a]}; eage[num]=e; head[a]=num++; } void tarjan(int u) { int v; low[u]=dfn[u]=++dfs_clock; instack[u]=1; s.push(u); for(int i=head[u];i!=-1;i=eage[i].next) { v=eage[i].to; if(!dfn[v]) { tarjan(v); low[u]=min(low[v],low[u]); } else if(instack[v]) low[u]=min(dfn[v],low[u]); } if(low[u]==dfn[u]) { scc_cnt++; scc[scc_cnt].clear(); do { v=s.top(); s.pop(); instack[v]=0; sccno[v]=scc_cnt; scc[scc_cnt].push_back(v); }while(u!=v); } } int in[MAXN],out[MAXN]; int main() { int n,m,a,b,x,i,j; while(scanf("%d%d",&n,&m),n) { memset(head,-1,sizeof(head)); num=0; while(m--) { scanf("%d%d",&a,&b); add(a,b); } scc_cnt=dfs_clock=0; memset(dfn,0,sizeof(dfn)); memset(low,0,sizeof(low)); memset(instack,0,sizeof(instack)); memset(sccno,0,sizeof(sccno)); for(i=1;i<=n;i++) if(!dfn[i]) tarjan(i); memset(in,0,sizeof(in)); for(i=1;i<=n;i++) { for(j=head[i];j!=-1;j=eage[j].next) if(sccno[i]!=sccno[eage[j].to]) { in[sccno[i]]=1; break; } } int sum=0; for(i=1;i<=scc_cnt;i++) if(!in[i]) { for(j=1;j<=n;j++) if(sccno[j]==i) out[sum++]=j; } sort(out,out+sum); if(sum!=0) { for(i=0;i<sum-1;i++) printf("%d ",out[i]); printf("%d\n",out[sum-1]); } else { printf("\n"); continue; } } return 0; }