题目地址：POJ 2762 先缩点，然后判断拓扑网络每层的个数是否为1（我承认如果事先不知道这题需要拓扑排序我是想不出来这点的。。。）。因为假如有一层为2的话，那么从此之后这两个岔路的点就不可能从一点到另一点的。 代码如下：
;mod=1e9+7;const int INF=0x3f3f3f3f;const double eqs=1e-9;const int MAXN=1000+10;int head[MAXN], Ecnt, scc, top, indx;int low[MAXN], dfn[MAXN], belong[MAXN], stk[MAXN], instack[MAXN];struct node{int u, v, next;}edge[7000];void add(int u, int v){edge[Ecnt].v=v;edge[Ecnt].next=head[u];head[u]=Ecnt++;}void tarjan(int u){low[u]=dfn[u]=++indx;instack[u]=1;stk[++top]=u;for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].v;if(!dfn[v]){tarjan(v);low[u]=min(low[u],low[v]);}else if(instack[v]){low[u]=min(low[u],dfn[v]);}}if(low[u]==dfn[u]){scc++;while(1){int v=stk[top–];belong[v]=scc;instack[v]=0;if(u==v) break;}}}void init(){memset(head,-1,sizeof(head));memset(dfn,0,sizeof(dfn));memset(instack,0,sizeof(instack));Ecnt=top=indx=scc=0;}int head1[MAXN], Ecnt1, deg[MAXN];struct node1{int u, v, next;}edge1[7000];void add1(int u, int v){edge1[Ecnt1].v=v;edge1[Ecnt1].next=head1[u];head1[u]=Ecnt1++;}bool topo(int scc){int i, u, tot, m=scc;while(m–){tot=0;for(i=1;i<=scc;i++){if(!deg[i]){tot++;u=i;deg[i]–;}}if(tot>1) return false;for(i=head1[u];i!=-1;i=edge1[i].next){deg[edge1[i].v]–;}}return true;}void init1(){memset(deg,0,sizeof(deg));memset(head1,-1,sizeof(head1));Ecnt1=0;}int main(){int n, m, i, j, u, v, t, f;scanf(“%d”,&t);while(t–){scanf(“%d%d”,&n,&m);init();while(m–){scanf(“%d%d”,&u,&v);add(u,v);}for(i=1;i<=n;i++){if(!dfn[i])tarjan(i);}init1();for(i=1;i<=n;i++){for(j=head[i];j!=-1;j=edge[j].next){v=edge[j].v;if(belong[i]!=belong[v]){add1(belong[i],belong[v]);deg[belong[v]]++;//printf(“%d %d\n”,belong[i],belong[v]);}}}if(topo(scc)) puts(“Yes”);else puts(“No”);}return 0;}
，人生伟业的建立，不在能知，乃在能行。
