题目请点我 题解： 这也是一道需要强连通提前处理的题目，但是我们需要的信息没有那么多。首先如果有满足解的话一定是所有的点都能指向终点（出度为0），并且重点只有1个，我们要求的答案也就是终点的size，，如果不符合上述情况就代表无解，输出0.所以我们只需要知道出度为0的点的个数以及每个节点的size就可以了。 代码实现：
;struct E{int from,to,next;};int N,M;int result;int edgenum;int top,Time,cnt;E edge[MAX_M];int dis[MAX_N];int head[MAX_N];int belong[MAX_N];bool instack[MAX_N];int outdegree[MAX_N];vector<int> G[MAX_N];vector<int> bcnt[MAX_N];int DFN[MAX_N],Low[MAX_N],Stack[MAX_N];int solve();void suodian();void tarjan(int u);void add(int u,int v);void tarjan_ini(int all);int main(){while( scanf(“%d%d”,&N,&M) != EOF ){result = 0;edgenum = 0;memset(head,-1,sizeof(head));while( M– ){int a,b;scanf(“%d%d”,&a,&b);add(a,b);}tarjan_ini(N);suodian();result = solve();printf(“%d\n”,result);}return 0;}void add(int u,int v){E tmp={u,v,head[u]};edge[edgenum] = tmp;head[u] = edgenum++;}void tarjan_ini(int all){memset(dis,0,sizeof(dis));memset(DFN,-1,sizeof(DFN));top = cnt = Time = 0;for( int i = 1; i <= all; i++ ){if( DFN[i] == -1 ){tarjan(i);}}}void tarjan(int u){DFN[u] = Low[u] = Time++;Stack[++top] = u;instack[u] = true;for( int i = head[u]; i != -1; i = edge[i].next ){int v = edge[i].to;if( DFN[v] == -1 ){tarjan(v);Low[u] = min(Low[u],Low[v]);}else if( instack[v] ){Low[u] = min(Low[u],DFN[v]);}}if( DFN[u] == Low[u] ){int now;int tmp = 0;cnt++;bcnt[cnt].clear();do{tmp++;now = Stack[top–];belong[now] = cnt;instack[now] = false;//bcnt[cnt].push_back(now);}while( now != u );dis[cnt] = tmp;}return ;}void suodian(){for( int i = 1; i <= cnt; i++ ){G[i].clear();outdegree[i] = 0;}for( int i = 0; i < edgenum; i++ ){int u,v;u = belong[edge[i].from];v = belong[edge[i].to];if( u != v ){//G[u].push_back(v);outdegree[u]++;}}}int solve(){int tmp = 0;int pos = 0;for( int i = 1; i <= cnt; i++ ){if( outdegree[i] == 0 ){tmp++;pos = i;}}if( tmp != 1 ){return 0;}else{//return bcnt[pos].size();return dis[pos];}}
环境不会改变，解决之道在于改变自己。
