poj2553--Tarjan

poj2553--Tarjan算法

题目大意：

    此题主要难点在于题目的理解：A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from v, v is also reachable from w.

sink 就是指某个能到达的点也能到达他自己的，即他们在同一个强联通分量中。

题目解法：

    找到出度为0的强联通分量中的所有点都是符合要求的。题目还要求将最终答案排序后输出，这里wa了一次，我以为是每个强联通分量单独排序。

题目源码：

//#define LOCAL
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;

#define MAXN 5500
#define INF 0x3f3f3f3f
int n, m;

struct Edge
{
	int to;
	int next;
};

Edge edge[MAXN * MAXN];
int head[MAXN], e;
int dfn[MAXN], low[MAXN], cnt;
int stack[MAXN], top;
int instack[MAXN];
int belong[MAXN], index;
int out[MAXN];
int ans[MAXN];

int min(int a, int b)
{
	return a < b ? a : b;
}

void addEdge(int u, int v)
{
	edge[e].to = v;
	edge[e].next = head[u];
	head[u] = e++;
}



void init()
{
	int i, j;
	int u, v;
	e = 0;
	cnt = 1;
	top = 0;
	index = 0;
	memset(head, -1, sizeof(head)); 
	memset(dfn, -1, sizeof(dfn));
	memset(low, -1, sizeof(low)); 
	memset(instack, 0, sizeof(instack));
	memset(belong, 0, sizeof(belong));
	memset(out, 0, sizeof(out)); 
	memset(ans, 0, sizeof(ans));
	for(i = 1; i <= m; i++)
	{
		scanf("%d%d", &u, &v);
		addEdge(u, v);
	} 
}

void tarjan(int u)
{
	int i, v;
	low[u] = dfn[u] = cnt++;
	instack[u] = 1;
	stack[++top] = u;
	for(i = head[u]; i != -1; i = edge[i].next)
	{
		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(low[u] == dfn[u])
	{
		index++;
		do
		{
			v = stack[top--];
			belong[v] = index;
		}
		while(top != 0 && v != u);
	}
}

void solve()
{
	int i, j;
	int v;
	int flag;
	while(scanf("%d%d", &n, &m) && n)
	{
		flag = 1;
		init(); // 获得head数组及edge数组 
		for(i = 1; i <= n; i++)
			if(dfn[i] == -1)
				tarjan(i); // 获得belong数组 
		for(i = 1; i <= n; i++)
		{
			for(j = head[i]; j != -1; j = edge[j].next) // 获得out数组 
			{
				v = edge[j].to;
				if(belong[i] != belong[v])
					out[belong[i]]++;
			}
			
		}
		flag = 1;
		
		for(i = 1; i <= n; i++)
		{
			if(!out[belong[i]])
			{
				if(flag)
				{
					printf("%d", i);
					flag = 0;
				}
				else
				{
					printf(" %d", i);
				}
			}
		}
		
		
		printf("\n");
	}
}

int main()
{
#ifdef LOCAL
	freopen("poj2553.txt", "r", stdin);
	freopen("poj2553Out.txt", "w", stdout);
#endif
	solve();
	return 0;
}