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hdu 1827 强连通缩点

2012-07-24 15:42 232 查看
#include <iostream>
#include <cstring>
#include <stack>
#define INF 0x3f3f3f3f
#define clr(x, k) memset((x),(k),sizeof(x))
using namespace std;

const int N = 1001;

struct edge
{
	int s, e, next;
}edg1[2*N], edg2[2*N];

int n, m, ct, index, et1, et2;
int head
, dfn
, low
, belong
;
int p
, indegree
, mmm
;
bool instack
;
stack <int> s;

void add_e(int u, int v, edge edg[], int &et)
{
	edg[et].s = u;
	edg[et].e = v;
	edg[et].next = head[u];
	head[u] = et++;
}

void read()
{
	int i, u, v;
	for (i=1;i<=n;++i)
	{
		scanf("%d", &p[i]);
	}
	et1 = 0;
	clr(head, -1);
	while (m--)
	{
		scanf("%d %d", &u, &v);
		add_e(u, v, edg1, et1);
	}
}

void tarjan(int i)
{
	int j;
	dfn[i] = low[i] = ++index;
	s.push(i);
	instack[i] = true;
	for (int u=head[i];u!=-1;u=edg1[u].next)
	{
		j = edg1[u].e;
		if (dfn[j]==0)
		{
			tarjan(j);
			if (low[i]>low[j]) low[i] = low[j];
		}
		else
		if (instack[j] && low[i]>dfn[j])
		    low[i] = dfn[j];
	}
	if (low[i] == dfn[i])
	{
		ct++;
		do
		{
			j = s.top(), s.pop();
			instack[j] = false;
			belong[j] = ct;
			if (mmm[ct]>p[j]) mmm[ct] = p[j];
		}while (i!=j);
	}
}

void ace()
{
	int i;
		clr(mmm, INF);
	ct = index = 0;
	clr(dfn, 0);
	clr(instack, 0);
	for (i=1;i<=n;++i)
	{
		if (!dfn[i]) tarjan(i);
	}
	
	int x;
	et2 = 0;
	clr(head, -1);
	clr(indegree, 0);

	for (i=0;i<et1;++i)
	{
		int j = edg1[i].s;
		int k = edg1[i].e;
		if (belong[j] != belong[k])
		{
			add_e(belong[j], belong[k], edg2, et2);
			indegree[belong[k]]++;
		}
//	  if (mmm[belong[j]]>p[j]) mmm[belong[j]] = p[j];
//	  if (mmm[belong[k]]>p[k]) mmm[belong[k]] = p[k];
	}
	int sum = 0, num = 0;
	for (i=1;i<=ct;i++)
	{
		if (indegree[i]==0)
		{
			sum += mmm[i];
			num++;
		}
	}
	printf("%d %d\n", num, sum);
}
			
int main()
{
	while (~scanf("%d %d", &n, &m))
	{
		read();
		ace();
	}
	return 0;
}

划掉的两个if:mmm数组要在tarjan里更新,可能有孤点存在。。。。

for simplicity 借鉴了鸟神的ace~
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