POJ-2186 Popular Cows 强连通分量
题目链接：http://poj.org/problem?id=2186
求出度为0的强连通分量的点的个数。
1 //STATUS:C++_AC_94MS_920KB 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=10010; 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 struct Edge{ 55 int u,v; 56 }e[N*5]; 57 int first[N],next[N*5],pre[N],sccno[N],low[N],vis[N],p[N]; 58 int n,m,mt,dfs_clock,scnt; 59 stack<int> s; 60 61 int find(int x){return p[x]==x?x:p[x]=find(p[x]);} 62 63 void adde(int a,int b) 64 { 65 e[mt].u=a;e[mt].v=b; 66 next[mt]=first[a],first[a]=mt++; 67 } 68 69 void dfs(int u) 70 { 71 int i,j,v; 72 pre[u]=low[u]=++dfs_clock; 73 s.push(u); 74 for(i=first[u];i!=-1;i=next[i]){ 75 v=e[i].v; 76 if(!pre[v]){ 77 dfs(v); 78 low[u]=Min(low[u],low[v]); 79 } 80 else if(!sccno[v]){ 81 low[u]=Min(low[u],low[v]); 82 } 83 } 84 if(low[u]==pre[u]){ 85 int x=-1; 86 scnt++; 87 while(x!=u){ 88 x=s.top();s.pop(); 89 sccno[x]=scnt; 90 } 91 } 92 } 93 94 int main() 95 { 96 // freopen("in.txt","r",stdin); 97 int i,j,a,b,ans,x,y,ok; 98 while(~scanf("%d%d",&n,&m)) 99 { 100 mem(first,-1);mt=0; 101 for(i=1;i<=n;i++)p[i]=i; 102 for(i=0;i<m;i++){ 103 scanf("%d%d",&a,&b); 104 x=find(a);y=find(b); 105 if(x!=y)p[y]=p[x]; 106 adde(a,b); 107 } 108 ok=0; 109 for(i=1;i<=n;i++){ 110 if(p[i]==i)ok++; 111 if(ok>=2)break; 112 } 113 ans=0; 114 if(ok==1){ 115 mem(pre,0);mem(sccno,0); 116 scnt=dfs_clock=0; 117 for(i=1;i<=n;i++){ 118 if(!pre[i])dfs(i); 119 } 120 for(i=1;i<=scnt;i++)vis[i]=1; 121 for(i=0;i<mt;i++){ 122 if(sccno[e[i].u]!=sccno[e[i].v]){ 123 vis[sccno[e[i].u]]=0; 124 } 125 } 126 ok=0; 127 for(i=1;i<=scnt;i++){ 128 ok+=vis[i]; 129 if(ok>=2){ok=0;break;} 130 } 131 if(ok){ 132 for(i=1;i<=n;i++){ 133 if(vis[sccno[i]])ans++; 134 } 135 } 136 } 137 138 printf("%d\n",ans); 139 } 140 return 0; 141 }
