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POJ 2553 taarjan缩点+度数
The Bottom of a Graph
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 8904 | Accepted: 3689 |
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.
Sample Input
3 31 3 2 3 3 12 11 20
Sample Output
1 32
题目意思:
给一个n个结点,m条边的有向图,若一个点所能到达的其他点都能到达该点或一个点不能到达其他点,那么该点为sink,从小到达输出sink。
思路:
这个题难在翻译上,特别是我这四级考了两次才过的人T-T,看懂了就很水了,先用tarjan缩点分块,然后求出没有出度的块的点标记下来,最后排序输出就行了。
代码:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <vector> 6 #include <queue> 7 #include <stack> 8 using namespace std; 9 10 #define N 500511 int n, m;12 vector<int>ve[N];13 int low[N], dfn[N], dfn_clock;14 int in[N], out[N], instack[N], a[N], b[N];15 int cnt;16 stack<int>st;17 18 void init(){19 int i, j;20 while(!st.empty()) st.pop();21 22 for(i=0;i<=n;i++){23 ve[i].clear();instack[i]=1;24 }25 memset(dfn,0,sizeof(dfn));26 memset(in,0,sizeof(in));27 memset(out,0,sizeof(out));28 }29 30 void tarjan(int u){31 int i, j, v;32 dfn[u]=low[u]=dfn_clock++;33 st.push(u);34 for(i=0;i<ve[u].size();i++){35 v=ve[u][i];36 if(!dfn[v]){37 tarjan(v);38 low[u]=min(low[u],low[v]);39 }40 else if(instack[v]){41 low[u]=min(low[u],dfn[v]);42 }43 }44 if(low[u]==dfn[u]){45 cnt++;46 while(1){47 v=st.top();st.pop();48 a[v]=cnt;instack[v]=0;49 if(v==u) break;50 }51 }52 }53 54 main()55 {56 int i, j, k;57 int x, y;58 while(scanf("%d",&n)==1&&n){59 scanf("%d",&m);60 init();61 while(m--){62 scanf("%d %d",&x,&y);63 ve[x].push_back(y);64 }65 dfn_clock=1;cnt=0;66 for(i=1;i<=n;i++){67 if(!dfn[i])68 tarjan(i);69 }70 for(i=1;i<=n;i++){71 for(j=0;j<ve[i].size();j++){72 x=ve[i][j];73 if(a[x]!=a[i]){74 out[a[i]]++;in[a[x]]++;75 }76 }77 }78 k=0;79 for(i=1;i<=n;i++){80 if(!out[a[i]]){81 b[k++]=i;82 }83 }84 sort(b,b+k);85 printf("%d",b[0]);86 for(i=1;i<k;i++) printf(" %d",b[i]);87 cout<<endl;88 }89 }
POJ 2553 taarjan缩点+度数
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