POJ2762 Going from u to v or from v to u? 双连通分量+拓扑排序

　　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 }