题目大意

　　求一个有向图含节点数最多且结点编号从小到大排列字典序最小的强连通分量。

注意事项

　　HDU1269那道题题面、数据太弱，在这道题上把我害惨了。。。

1. Dfs点u时，如果与u相连的一个点v有DfsN，v必须在栈内我们才能更新u的Low值。
2. 我们要枚举每个点作为Dfs的起始点，不能只枚举一个。
3. Dfs之前不必要把所有结点的标记清空，因为打过标记的点不会与后来Dfs到的点形成强连通分量，而且会导致重复计算。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>
using namespace std;const int MAX_NODE = 5010, INF = 0x3f3f3f3f;struct Node
{
vector<Node*> Next;
int DfsN, Low;
bool InStack;
}_nodes[MAX_NODE];
int TotNode, TotEdge;
int AnsConIds[MAX_NODE], TempConIds[MAX_NODE];//Con::Connect
int AnsConCnt, TempConCnt;
int CurDfsN;
stack<Node*> St;void AddEdge(int u, int v, int k)
{
_nodes[u].Next.push_back(_nodes + v);
if (k == 2)
_nodes[v].Next.push_back(_nodes + u);
}bool CmpArg(int *a, int *b, int n)
{
for (int i = 1; i <= n; i++)
if (a[i] != b[i])
return a[i] < b[i];
return false;
}void UpdateAns()
{
if (TempConCnt >= AnsConCnt)
{
sort(TempConIds + 1, TempConIds + TempConCnt + 1);
if (TempConCnt > AnsConCnt || CmpArg(TempConIds, AnsConIds, AnsConCnt))
{
memcpy(AnsConIds, TempConIds, sizeof(TempConIds));
AnsConCnt = TempConCnt;
}
}
}void PopStack(Node *cur)
{
TempConCnt = 0;
Node *temp;
do {
temp = St.top();
temp->InStack = false;
TempConIds[++TempConCnt] = temp - _nodes;
St.pop();
} while (temp != cur);
UpdateAns();
}void Dfs(Node *cur)
{
cur->DfsN = ++CurDfsN;
cur->Low = cur->DfsN;
St.push(cur);
cur->InStack = true;
int nextCnt = cur->Next.size();
for (int i = 0; i < nextCnt; i++)
{
if (!cur->Next[i]->DfsN)
{
Dfs(cur->Next[i]);
cur->Low = min(cur->Low, cur->Next[i]->Low);
}
else if(cur->Next[i]->InStack)
cur->Low = min(cur->Low, cur->Next[i]->DfsN);
}
if (cur->Low == cur->DfsN)
PopStack(cur);
}int main()
{
scanf("%d%d", &TotNode, &TotEdge);
for (int i = 1; i <= TotEdge; i++)
{
int u, v, k;
scanf("%d%d%d", &u, &v, &k);
AddEdge(u, v, k);
}
CurDfsN = 0;
for (int i = 1; i <= TotNode; i++)
{
if (!_nodes[i].DfsN)
{
_printf("from %d:\n", i);
while (St.size())
St.pop();
CurDfsN = 0;
Dfs(_nodes + i);
}
}
printf("%d\n", AnsConCnt);
sort(AnsConIds + 1, AnsConIds + AnsConCnt + 1);
for (int i = 1; i <= AnsConCnt; i++)
printf("%d ", AnsConIds[i]);
printf("\n");
return 0;
}