HDU 1269 迷宫城堡 scc

判断强联通scc是否为1

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN = 20010;//点数 
 4 const int MAXM = 200100;//边数 
 5 struct Edge {  
 6     int to,next; 
 7 }edge[MAXM]; 
 8 int head[MAXN],tot; 
 9 int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];//Belong数组的值是1~scc 
10 int Index,top; int scc;//强连通分量的个数 
11 
12 bool Instack[MAXN]; 
13 int num[MAXN];//各个强连通分量包含点的个数,数组编号1~scc //num数组不一定需要,结合实际情况 
14  
15 void addedge(int u,int v){  
16     edge[tot].to = v;
17     edge[tot].next = head[u];
18     head[u] = tot++; 
19 } 
20 void Tarjan(int u){
21     int v;  
22     Low[u] = DFN[u] = ++Index; 
23     Stack[top++] = u;  
24     Instack[u] = true;  
25     for(int i = head[u];i != -1;i = edge[i].next) {  
26         v = edge[i].to;   
27         if( !DFN[v] ){
28             Tarjan(v);   
29             if( Low[u] > Low[v] )
30                 Low[u] = Low[v];   
31         }   
32         else if(Instack[v] && Low[u] > DFN[v])
33             Low[u] = DFN[v];  
34         }  
35         if(Low[u] == DFN[u]){ 
36             scc++; 
37             do{    
38                 v = Stack[--top];
39                 Instack[v] = false;    
40                 Belong[v] = scc;    
41                 num[scc]++;   
42             }while( v != u);  
43         } 
44 } 
45 void solve(int N){
46     memset(DFN,0,sizeof(DFN));
47     memset(Instack,false,sizeof(Instack));
48     memset(num,0,sizeof(num));  
49     Index = scc = top = 0;  
50     for(int i = 1;i <= N;i++)
51         if(!DFN[i])    
52         Tarjan(i); 
53 } 
54 void init(){
55     tot = 0; 
56     memset(head,-1,sizeof(head)); 
57 } 
58 int main(){
59     int n,m,u,v;
60     while(scanf("%d%d",&n,&m)!=EOF){
61         if(!n&&!m) break;
62         init();
63         for(int i=1;i<=m;i++){
64             scanf("%d%d",&u,&v);
65             addedge(u,v);
66         }
67         solve(n);
68         bool flag=true;
69 //        cout<<scc<<endl; 
70         if(scc==1) cout<<"Yes"<<endl;
71         else cout<<"No"<<endl;
72     }
73     return 0;
74 }

 

posted on 2017-08-18 12:26  poler  阅读(163)  评论(0编辑  收藏  举报

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