{"id":377,"date":"2013-03-31T17:17:00","date_gmt":"2013-03-31T09:17:00","guid":{"rendered":"http:\/\/nike0good.jp1.rpvhost.net\/281"},"modified":"2013-03-31T17:17:00","modified_gmt":"2013-03-31T09:17:00","slug":"poj_2553_the_bottom_of_a_graph-reduction_of_degrees_obtained","status":"publish","type":"post","link":"https:\/\/nike0good.com\/?p=377","title":{"rendered":"POJ 2553(The Bottom of a Graph-\u7f29\u70b9\u6c42\u51fa\u5ea6)"},"content":{"rendered":"\n\n\n
\n
Language:\n\n\n\n
Time Limit:<\/strong> 3000MS<\/td>\n <\/td>\nMemory Limit:<\/strong> 65536K<\/td>\n<\/tr>\n
Total Submissions:<\/strong> 7308<\/td>\n <\/td>\nAccepted:<\/strong> 3003<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Description<\/p>\n

\nWe will use the following (standard) definitions from graph theory. Let V<\/em> be a nonempty and finite set, its elements being called vertices (or nodes). Let E<\/em> be a subset of the Cartesian product V\u00d7V<\/em>, its elements being called edges.
\n Then G=(V,E)<\/em> is called a directed graph. 
\nLet n<\/em> be a positive integer, and let p=(e1<\/sub>,...,en<\/sub>)<\/em> be a sequence of length n<\/em> of edges ei<\/sub>\u2208E<\/em> such that ei<\/sub>=(vi<\/sub>,vi+1<\/sub>)<\/em> for a sequence of vertices (v1<\/sub>,...,vn+1<\/sub>)<\/em>.
\n Then p<\/em> is called a path from vertex v1<\/sub><\/em> to vertex vn+1<\/sub><\/em> in G<\/em> and we say that vn+1<\/sub><\/em> is reachable from v1<\/sub><\/em>, writing (v1<\/sub>\u2192vn+1<\/sub>)<\/em>. 
\nHere are some new definitions. A node v<\/em> in a graph G=(V,E)<\/em> is called a sink, if for every node w<\/em> in G<\/em> that is reachable from v<\/em>, v<\/em> is also reachable from w<\/em>. The bottom of a graph is the subset of
\n all nodes that are sinks, i.e.,bottom(G)={v\u2208V|\u2200w\u2208V:(v\u2192w)\u21d2(w\u2192v)}<\/em>. You have to calculate the bottom of certain graphs.<\/div>\n

Input<\/p>\n

\nThe input contains several test cases, each of which corresponds to a directed graph G<\/em>. Each test case starts with an integer number v<\/em>, denoting the number of vertices of G=(V,E)<\/em>, where the vertices will be identified by the integer
\n numbers in the set V={1,...,v}<\/em>. You may assume that 1<=v<=5000<\/em>. That is followed by a non-negative integer e<\/em> and, thereafter, e<\/em> pairs of vertex identifiers v1<\/sub>,w1<\/sub>,...,ve<\/sub>,we<\/sub><\/em> with
\n the meaning that (vi<\/sub>,wi<\/sub>)\u2208E<\/em>. There are no edges other than specified by these pairs. The last test case is followed by a zero.<\/div>\n

Output<\/p>\n

\nFor 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.\"\"<\/div>\n

Sample Input<\/p>\n

3 3\n1 3 2 3 3 1\n2 1\n1 2\n0\n<\/pre>\n
Sample Output<\/p>\n
1 3\n2\n<\/pre>\n
Source<\/p>\n
\nUlm Local 2003<\/a><\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
<\/p>\n
Tarjen\u7f29\u70b9\u7edf\u8ba1\u51fa\u5ea6\u3002<\/span><\/p>\n

\n<\/span><\/p>\n
<\/span><\/p>\n
#include<cstdio>\n#include<cstdlib>\n#include<algorithm>\n#include<functional>\n#include<cstring>\n#include<iostream>\nusing namespace std;\n#define MAXN (5000+10)\n#define MAXM (1000000+10)\nint n,m,h[MAXN],t[MAXN],dfs[MAXN],tim,kind,s[MAXN],tot;\nbool b[MAXN];\nint edge[MAXM],next[MAXM],pre[MAXN],size;\nint x[MAXM],y[MAXM],numk[MAXN];\nvoid addedge(int u,int v)\n{\n\tedge[++size]=v;\n\tnext[size]=pre[u];\n\tpre[u]=size;\n}\nvoid tar(int u)\n{\n\tdfs[u]=t[u]=++tim;b[u]=1;s[++tot]=u;\n\tfor (int p=pre[u];p;p=next[p])\n\t{\n\t\tint &v=edge[p];\n\t\tif (!b[v]) tar(v),dfs[u]=min(dfs[u],dfs[v]);\n\t\telse if (!h[v]) dfs[u]=min(dfs[u],t[v]);  \/\/\u4fdd\u8bc1\u6307\u5411\u7956\u5148\n\t}\n\tif (dfs[u]==t[u])\n\t{\n\t\tkind++;\n\t\twhile (s[tot]!=u) h[s[tot--]]=kind,numk[kind]++;\n\t\th[s[tot--]]=kind;numk[kind]++;\n\n\t}\n}\nint outdegree[MAXN];\nint main()\n{\n\twhile (scanf("%d",&n)&&n)\n\t{\n\t\tscanf("%d",&m);\n\t\ttot=size=tim=kind=0;\n\t\tmemset(h,0,sizeof(h));\n\t\tmemset(t,0,sizeof(t));\n\t\tmemset(dfs,0,sizeof(dfs));\n\t\tmemset(s,0,sizeof(s));\n\t\tmemset(pre,0,sizeof(pre));\n\t\tmemset(b,0,sizeof(b));\n\t\tmemset(numk,0,sizeof(numk));\n\t\tmemset(outdegree,0,sizeof(outdegree));\n\n\t\tfor (int i=1;i<=m;i++)\n\t\t{\n\t\t\tscanf("%d%d",&x[i],&y[i]);\n\t\t\taddedge(x[i],y[i]);\n\t\t}\n\t\tfor (int i=1;i<=n;i++) if (!b[i]) tar(i);\n\/\/\t\tfor (int i=1;i<=n;i++) cout<<h[i]<<' ';cout<<endl;\n\t\tfor (int i=1;i<=m;i++)\n\t\t{\n\t\t\tif (h[x[i]]^h[y[i]]) outdegree[h[x[i]]]++;\n\t\t}\n\t\t\/*\n\t\tint ans=0;\n\t\tfor (int i=1;i<=kind;i++)\n\t\t\tif (!outdegree[i]) ans+=numk[i];\n\t\tcout<<ans<<endl;\n\t\t*\/\n\t\tbool flag=0;\n\t\tfor (int i=1;i<=n;i++)\n\t\t\tif (!outdegree[h[i]])\n\t\t\t{\n\t\t\t\tif (flag) printf(" ");else flag=1;printf("%d",i);\n\t\t\t}\n\n\n\t\/\/\tcout<<ans<<endl;\n\t\tcout<<endl;\n\t}\n\treturn 0;\n}\n<\/pre>\n<\/p>\n
<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"
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