The Bottom of a Graph

Time Limit: 3000ms

Memory Limit: 65536KB

This problem will be judged on 

PKU. Original ID: 

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

 

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 
        
          2 #include 
         
           3 #include 
          
            4 #include 
           
             5 #include 
            
              6 #include 
             
               7 #include 
              
                8 #include 
               
                 9 #include 
                
                 10 #include 
                 
                  11 #include 
                  
                   12 #include 
                   
                    13 #define LL long long14 #define pii pair
                    
                     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 }