POJ-2553 The Bottom of a Graph 强连通分量
题目链接：http://poj.org/problem?id=2553
题意：在有向图中，求出一些点，自己能到达的点一定能到达自己。
简化模型，就是求出度为0的强连通分量中的那些点。
1 //STATUS:C++_AC_79MS_712KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //define 25 #define pii pair<int,int> 26 #define mem(a,b) memset(a,b,sizeof(a)) 27 #define lson l,mid,rt<<1 28 #define rson mid+1,r,rt<<1|1 29 #define PI acos(-1.0) 30 //typedef 31 typedef __int64 LL; 32 typedef unsigned __int64 ULL; 33 //const 34 const int N=5010; 35 const int INF=0x3f3f3f3f; 36 const int MOD=100000,STA=8000010; 37 const LL LNF=1LL<<60; 38 const double EPS=1e-8; 39 const double OO=1e15; 40 const int dx[4]={-1,0,1,0}; 41 const int dy[4]={0,1,0,-1}; 42 //Daily Use ... 43 inline int sign(double x){return (x>EPS)-(x<-EPS);} 44 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 45 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 46 template<class T> inline T Min(T a,T b){return a<b?a:b;} 47 template<class T> inline T Max(T a,T b){return a>b?a:b;} 48 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 49 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 50 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 51 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 52 //End 53 54 int pre[N],low[N],sccno[N],ans[N],vis[N]; 55 int n,m,dfs_clock,k,scnt; 56 vector<int> g[N]; 57 stack<int> s; 58 59 int dfs(int u) 60 { 61 pre[u]=++dfs_clock; 62 int i,v,ok=1; 63 low[u]=pre[u]; 64 s.push(u); 65 for(i=0;i<g[u].size();i++){ 66 v=g[u][i]; 67 if(!pre[v]){ 68 dfs(v); 69 low[u]=Min(low[u],low[v]); 70 } 71 else if(!sccno[v]){ 72 low[u]=Min(low[u],low[v]); 73 } 74 } 75 if(low[u]==pre[u]){ 76 scnt++; 77 int x=-1; 78 while(x!=u){ 79 x=s.top(),s.pop(); 80 sccno[x]=scnt; 81 } 82 // printf(" %d %d %d\n",u,k,ok); 83 } 84 return ok; 85 } 86 87 int main() 88 { 89 // freopen("in.txt","r",stdin); 90 int i,j,a,b,ok; 91 while(~scanf("%d%d",&n,&m) && n) 92 { 93 for(i=1;i<=n;i++)g[i].clear(); 94 for(i=0;i<m;i++){ 95 scanf("%d%d",&a,&b); 96 g[a].push_back(b); 97 } 98 99 mem(pre,0);mem(sccno,0); 100 dfs_clock=scnt=0; 101 for(i=1;i<=n;i++){ 102 if(!pre[i])dfs(i); 103 } 104 for(i=1;i<=scnt;i++)vis[i]=1; 105 for(i=1;i<=n;i++){ 106 for(j=0;j<g[i].size();j++){ 107 if(sccno[g[i][j]]!=sccno[i]){ 108 vis[sccno[i]]=0; 109 break; 110 } 111 } 112 } 113 k=0; 114 for(i=1;i<=n;i++){ 115 if(vis[sccno[i]])ans[k++]=i; 116 } 117 sort(ans,ans+k); 118 printf("%d",ans[0]); 119 for(i=1;i<k;i++) 120 printf(" %d",ans[i]); 121 putchar('\n'); 122 } 123 return 0; 124 }
