HDU 3836 tarjan求强连通分量

2014-11-24 02:24:29 · 作者: · 浏览: 1
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define N 20200
#define inf 10000000
using namespace std;
inline int Min(int a,int b){return a>b b:a;}
inline int Max(int a,int b){return a 这是反向弧
			Low[u] = Min(Low[u], DFN[v]);
	}
	if(DFN[u] == Low[u])//反向弧在u及u以下
	{
		taj++;
		while(1){
			int now = Stack[--top];
			vis[now] = 0;
			belong[now] = taj;
			if(now == u)break;
		}
	}
}

void tarjan_init(){
		memset(head, -1, sizeof(head)); edgenum = 0;
		memset(DFN, -1, sizeof(DFN));
		memset(Low, -1, sizeof(Low));
		memset(vis, 0, sizeof(vis));
		taj = top = 0;
		Time = 0;
}
int n, m;
int outde[N], inde[N];

int main(){
	int i,j,u,v;
	while(~scanf("%d %d",&n,&m)){
		tarjan_init();

		while(m--){
			scanf("%d %d",&u,&v);
			addedge(u, v);
		}

		for(i=1;i<=n;i++)if(DFN[i]==-1)tarjan(i);
		if(taj == 1 || n<1){printf("0\n");continue;}

		int ans1 = 0, ans2 = 0;
		memset(inde, 0, sizeof(inde));
		memset(outde,0, sizeof(outde));
		for(i = 0;i