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UVA 11770 — Lighting Away 强连通+拓扑

2015-05-04 19:53 267 查看
题意:有n盏灯,m个关联;

a点亮了b必定点亮;

问至少要开几盏灯才能使所有的灯点亮;

思路:用强连通缩点,然后拓扑求入度为0得点;

#include<stdio.h>
#include<vector>
#include<string.h>
#include<iostream>
#include<algorithm>

using namespace std;
const int N = 10005;
int head
, vis
;
int DFN
, low
, belong
, stack
;
int n, m, edgenum;
int top, Time, taj;

struct node
{
	int from, to, nex;
	bool sign;
}edge[100005];

void add(int u, int v)
{
	edge[edgenum].from = u;
	edge[edgenum].to = v;
	edge[edgenum].sign = false;
	edge[edgenum].nex = head[u];
	head[u] = edgenum++;
}

vector<int>bcc
;
void tarjan(int u, int fa)
{
	DFN[u] = low[u] = Time++;
	vis[u] = 1;
	stack[top++] = u;
	for(int i = head[u];i!=-1;i = edge[i].nex)
	{
		int v = edge[i].to;
		if(DFN[v] == -1)
		{
			tarjan(v, u);
			low[u] = min(low[u], low[v]);
			if(DFN[u]<low[v])
				edge[i].sign = 1;
		}
		else if(vis[v])
		low[u] = min(low[u], DFN[v]);
	}
	if(low[u] == DFN[u])
	{  
		taj ++ ; 
		bcc[taj].clear();
		while(1)
		{
			int now = stack[--top];  
			vis[now] = 0; 
			belong[now] = taj;
			bcc[taj].push_back(now);
			if(now == u)
			break;
		}
	}
}

vector<int>G
;
int du
;
void suodian()
{
	memset(du, 0, sizeof(du));
	for(int i = 1;i<=taj;i++)
	G[i].clear();
	for(int i = 0;i<edgenum;i++)
	{
		int u = belong[edge[i].from];
		int v = belong[edge[i].to];
		if(u!=v)
		{
			G[u].push_back(v);
			du[v]++;
		}
	}
}

void tarjan_init(int all)
{
	memset(DFN, -1, sizeof(DFN));
	memset(low, -1, sizeof(low));
	memset(vis, 0, sizeof(vis));
	Time = top = taj = 0;
	for(int i = 1;i<=all;i++)
	{
		if(DFN[i] == -1)
		tarjan(i, i);
	}
}

int main()
{
	int cas;
	int t = 0;
	scanf("%d", &cas);
	while(cas--)
	{
		memset(head, -1, sizeof(head));
		edgenum = 0;
		scanf("%d%d", &n, &m);
		while(m--)
		{
			int u, v;
			scanf("%d%d", &u, &v);
			add(u, v);
		}
		tarjan_init(n);
		suodian();
		int con = 0;
		for(int i = 1;i<=taj;i++)
		{
			if(du[i] == 0)
			con++;
		}
		printf("Case %d: %d\n", ++t, con);
	}
	return 0;
}
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