POJ-2762 Going from u to v or from v to u? 双连通分量+拓扑排序
题目链接：http://poj.org/problem?id=2762
判断在一个有向图中，是否任意的两点存在一条通路。
首先用tarjan算法进行边-双连通分量缩点，接下来就是判断树的分支只有一个，那么就用拓扑排序每次判断入度为0的点是否只有一个。
1 //STATUS:C++_AC_360MS_340KB 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=1010; 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*6],e2[N*6]; 57 int first[N],next[N*6],first2[N],next2[N*6],pre[N],sccno[N],low[N],c[N],vis[N],p[N]; 58 int T,n,m,mt,dfs_clock,scnt,mt2; 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 adde2(int a,int b) 70 { 71 e2[mt2].u=a;e2[mt2].v=b; 72 next2[mt2]=first2[a],first2[a]=mt2++; 73 } 74 75 void dfs(int u) 76 { 77 int i,j,v; 78 pre[u]=low[u]=++dfs_clock; 79 s.push(u); 80 for(i=first[u];i!=-1;i=next[i]){ 81 v=e[i].v; 82 if(!pre[v]){ 83 dfs(v); 84 low[u]=Min(low[u],low[v]); 85 } 86 else if(!sccno[v]){ 87 low[u]=Min(low[u],low[v]); 88 } 89 } 90 if(low[u]==pre[u]){ 91 int x=-1; 92 scnt++; 93 while(x!=u){ 94 x=s.top();s.pop(); 95 sccno[x]=scnt; 96 } 97 } 98 } 99 100 int topo() 101 { 102 int i,j,cnt,sum=scnt,w; 103 mem(c,0); 104 for(;sum;sum--){ 105 mem(vis,0); 106 for(i=1;i<=scnt;i++){ 107 for(j=first2[i];j!=-1;j=next2[j]){ 108 vis[e2[j].v]=1; 109 } 110 } 111 cnt=0; 112 for(i=1;i<=scnt;i++){ 113 if(!c[i] && !vis[i]){w=i;cnt++;} 114 if(cnt>=2)return 0; 115 } 116 first2[w]=-1; 117 c[w]=1; 118 } 119 return 1; 120 } 121 122 int main() 123 { 124 // freopen("in.txt","r",stdin); 125 int i,j,a,b,x,y,ok; 126 scanf("%d",&T); 127 while(T--) 128 { 129 scanf("%d%d",&n,&m); 130 mem(first,-1);mt=0; 131 for(i=1;i<=n;i++)p[i]=i; 132 for(i=0;i<m;i++){ 133 scanf("%d%d",&a,&b); 134 x=find(a);y=find(b); 135 if(x!=y)p[y]=p[x]; 136 adde(a,b); 137 } 138 ok=0; 139 for(i=1;i<=n;i++){ 140 if(p[i]==i)ok++; 141 if(ok>=2){ok=0;break;} 142 } 143 if(ok==1){ 144 mem(pre,0);mem(sccno,0); 145 scnt=dfs_clock=0; 146 for(i=1;i<=n;i++){ 147 if(!pre[i])dfs(i); 148 } 149 if(scnt!=1){ 150 mt2=0; 151 mem(first2,-1); 152 for(i=0;i<mt;i++){ 153 if(sccno[e[i].u]!=sccno[e[i].v]){ 154 adde2(sccno[e[i].u],sccno[e[i].v]); 155 } 156 } 157 ok=topo(); 158 } 159 } 160 161 printf("%s\n",ok?"Yes":"No"); 162 } 163 return 0; 164 }