首页 > 代码库 > POJ 2553 The Bottom of a Graph

POJ 2553 The Bottom of a Graph

2024-08-02 00:29:31   222人阅读

The Bottom of a Graph

Time Limit: 3000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 2553
64-bit integer IO format: %lld      Java class name: Main
 
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.
 

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, epairs 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

Source

Ulm Local 2003
 
解题：求这样的点，它能到的点，那点也可以到它，注意先后顺序。然后求强联通缩点，出度为0的集合，里面的点即为我们求得那些点
 
 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib>10 #include <string>11 #include <set>12 #include <stack>13 #define LL long long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 using namespace std;17 const int maxn = 6000;18 struct arc {19     int to,next;20     arc(int x = 0,int y = -1) {21         to = x;22         next = y;23     }24 };25 arc e[maxn*100];26 int head[maxn],dfn[maxn],low[maxn],belong[maxn],my[maxn];27 int tot,n,m,top,scc,idx,out[maxn],ans[maxn];28 bool instack[maxn];29 void init() {30     for(int i = 0; i < maxn; ++i) {31         dfn[i] = low[i] = belong[i] = 0;32         instack[i] = false;33         head[i] = -1;34         out[i] = 0;35     }36     scc = tot = idx = top = 0;37 }38 void add(int u,int v){39     e[tot] = arc(v,head[u]);40     head[u] = tot++;41 }42 void tarjan(int u) {43     dfn[u] = low[u] = ++idx;44     my[top++] = u;45     instack[u] = true;46     for(int i = head[u]; ~i; i = e[i].next) {47         if(!dfn[e[i].to]) {48             tarjan(e[i].to);49             low[u] = min(low[u],low[e[i].to]);50         } else if(instack[e[i].to]) low[u] = min(low[u],dfn[e[i].to]);51     }52     if(dfn[u] == low[u]) {53         scc++;54         int v;55         do {56             v = my[--top];57             instack[v] = false;58             belong[v] = scc;59         } while(v != u);60     }61 }62 int main() {63     int u,v;64     while(~scanf("%d",&n)&&n) {65         scanf("%d",&m);66         init();67         for(int i = 0; i < m; ++i) {68             scanf("%d%d",&u,&v);69             add(u,v);70         }71         for(int i = 1; i <= n; ++i)72             if(!dfn[i]) tarjan(i);73         for(int i = 1; i <= n; ++i)74             for(int j = head[i]; ~j; j = e[j].next)75                 if(belong[i] != belong[e[j].to]) out[belong[i]]++;76         int cnt = 0;77         for(int i = 1; i <= n; ++i)78             if(!out[belong[i]]) ans[cnt++] = i;79         if(cnt){80             for(int i = 0; i < cnt; ++i)81                 printf("%d%c",ans[i],i + 1 == cnt?‘\n‘:‘ ‘);82         }else puts("");83     }84     return 0;85 }
View Code

 

POJ 2553 The Bottom of a Graph


联系
我们