hdu 1269（Tarjan求强连通分量）

这道题目就是求强连通分量。。。

这里采用的是Tarjan算法：http://m.blog.csdn.net/blog/qq574857122/16361033

AC代码：

#include<iostream>v
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 10005;
struct Edge
{
	int key,next;
}edge[100000];
int pre[maxn],cnt,n,m,dfsNum,sccNum,top;
int stack[maxn],instack[maxn],dfsnum[maxn],low[maxn];

void addedge(int x,int y)
{
	edge[cnt].key = y;
	edge[cnt].next = pre[x];
	pre[x] = cnt++;
}

void Tarjan(int cur)
{
	dfsnum[cur] = ++dfsNum;
	low[cur] = dfsnum[cur];
	stack[top++] = cur;
	instack[cur] = 1;
	for(int i = pre[cur]; i != -1; i = edge[i].next)
	{
		int v = edge[i].key;
		if(dfsnum[v] == 0) // 子树中还有边未遍历
		{
			Tarjan(v);
			if(low[cur] > low[v])
				low[cur] = low[v];
		}
		else if(instack[v])  //表示有父子关系，存在回边
		{
			if(low[cur] > dfsnum[v])
				low[cur] = dfsnum[v];
		}
	}
	if(low[cur] == dfsnum[cur])  //该点是强连通分量的根节点
	{
		while(top != 0 && stack[top-1] != cur)
		{
			instack[stack[top-1]] = 0;
			top--;
		}
		sccNum++;
	}
}

int main()
{	
	while(scanf("%d%d",&n,&m)!=EOF && m + n)
	{
		memset(pre,-1,sizeof(pre));
		memset(instack,0,sizeof(instack));
		cnt = top = sccNum = dfsNum = 0;
		for(int i = 1; i <= m; i++)
		{
			int a,b;
			scanf("%d%d",&a,&b);
			addedge(a,b);
		}
		for(int i = 1; i <= n; i++)
			if(dfsnum[i] == 0)
				Tarjan(i);
		if(sccNum == 1)
			printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}