hdoj 2767 Proving Equivalences 【有向图 增加最少的边使图强连通】



Proving Equivalences

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3897    Accepted Submission(s): 1409

Problem Description

Consider the following exercise, found in a generic linear algebra textbook.

Let A be an n × n matrix. Prove that the following statements are equivalent:

1. A is invertible.
2. Ax = b has exactly one solution for every n × 1 matrix b.
3. Ax = b is consistent for every n × 1 matrix b.
4. Ax = 0 has only the trivial solution x = 0.

The typical way to solve such an exercise is to show a series of implications. For instance, one can proceed by showing that (a) implies (b), that (b) implies (c), that (c) implies (d), and finally that (d) implies (a). These four implications show that the four statements are equivalent.

Another way would be to show that (a) is equivalent to (b) (by proving that (a) implies (b) and that (b) implies (a)), that (b) is equivalent to (c), and that (c) is equivalent to (d). However, this way requires proving six implications, which is clearly a lot more work than just proving four implications!

I have been given some similar tasks, and have already started proving some implications. Now I wonder, how many more implications do I have to prove? Can you help me determine this?

 

Input

On the first line one positive number: the number of testcases, at most 100. After that per testcase:

* One line containing two integers n (1 ≤ n ≤ 20000) and m (0 ≤ m ≤ 50000): the number of statements and the number of implications that have already been proved.
* m lines with two integers s1 and s2 (1 ≤ s1, s2 ≤ n and s1 ≠ s2) each, indicating that it has been proved that statement s1 implies statement s2.

 

Output

Per testcase:

* One line with the minimum number of additional implications that need to be proved in order to prove that all statements are equivalent.

 

Sample Input

2
4 0
3 2
1 2
1 3

 

Sample Output

4
2

 

等价命题：a <=> b  即 a可以推出b 且 b可以推出a。

题意：有n个命题，现已给出m次推导即由a可以推出b（根据该条件b不能推出a），问最少还需要多少次推导才可以证明n个命题之间是等价的。

分析   --   把每个命题虚拟成一个节点，a推导b当作一条a到b的有向边，问题就变成：在有向图中最少增加多少条边才可以使新图强连通。

思路：求出所有SCC，缩点并求出对应SCC的入度和出度。这时假设原图成为一个有K个点的有向图，我们的目的就是让这个新图强连通。

现在问题来了，对于每个节点（每个缩点后的SCC）它会有不同的出度入度，我们该怎么处理？

方案：求出所有入度为0的点数sumin，求出所有出度为0的点数sumout，取两者中较大值即可。

邻接表 爆到死边数太多了吧，改成容器vector就AC了，郁闷啊！！！

AC代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#define MAXN 20000+10
#define MAXM 400000000+10
#define INF 10000000
using namespace std;
//struct Edge
//{
//	int from, to, next;
//}edge[MAXN];
//int head[MAXN], edgenum;
vector<int> G[MAXN];//已推导的公式建立的图 
int low[MAXN], dfn[MAXN];
int dfs_clock;
int sccno[MAXN], scc_cnt;
vector<int> scc[MAXN];//存储SCC的点 
stack<int> S;
bool Instack[MAXN];
int n, m;//n个命题 m次推导
//void init()
//{
//	edgenum = 0;
//	memset(head, -1, sizeof(head));
//}
//void addEdge(int u, int v)
//{
//	Edge E = {u, v, head[u]};
//	edge[edgenum] = E;
//	head[u] = edgenum++;
//}
void getMap()
{
	int s, t;
	for(int i = 1; i <= n; i++) G[i].clear(); 
	while(m--)
	{
		scanf("%d%d", &s, &t);
		G[s].push_back(t);
	} 
}
void tarjan(int u, int fa)
{
	int v;
	low[u] = dfn[u] = ++dfs_clock;
	S.push(u);
	Instack[u] = true;
	for(int i = 0; i < G[u].size(); i++)
	{
		v = G[u][i];
		if(!dfn[v])
		{
			tarjan(v, u);
			low[u] = min(low[u], low[v]);
		} 
		else if(Instack[v])
		low[u] = min(low[u], dfn[v]);
	}
	if(low[u] == dfn[u])
	{
		scc_cnt++;
		scc[scc_cnt].clear();
		for(;;)
		{
			v = S.top(); S.pop();
			Instack[v] = false;
			sccno[v] = scc_cnt;
			scc[scc_cnt].push_back(v);
			if(v == u) break;
		}
	} 
}
void find_cut(int l, int r)
{
	memset(low, 0, sizeof(low));
	memset(dfn, 0, sizeof(dfn));
	memset(sccno, 0, sizeof(sccno));
	memset(Instack, false, sizeof(Instack));
	dfs_clock = scc_cnt = 0;
	for(int i = l; i <= r; i++)
	if(!dfn[i]) tarjan(i, -1); 
}
int in[MAXN], out[MAXN];//SCC入度 出度 
void suodian()//缩点 
{
	for(int i = 1; i <= scc_cnt; i++) in[i] = out[i] = 0; 
	for(int i = 1; i <= n; i++)
	{
		for(int j = 0; j < G[i].size(); j++)
		{
			int u = sccno[i];
			int v = sccno[G[i][j]];
			if(u != v)
			out[u]++, in[v]++;
		} 
	} 
//	for(int i = 0; i < edgenum; i++)
//	{
//		int u = sccno[edge[i].from];
//		int v = sccno[edge[i].to];
//		if(u != v)
//		G[u].push_back(v), out[u]++, in[v]++;
//	} 
}
void solve()
{
	if(scc_cnt == 1)
	{
		printf("0\n");
		return ;
	}
	int sumin = 0, sumout = 0;//入度为0的点数 出度为0的点数
	for(int i = 1; i <= scc_cnt; i++)
	{
		if(in[i] == 0) sumin++;
		if(out[i] == 0) sumout++;
	} 
	printf("%d\n", max(sumin, sumout));
} 
int main()
{
	int t;
	scanf("%d", &t);
	while(t--)
	{
		scanf("%d%d", &n, &m);
		//init();
		getMap();
		find_cut(1, n);//找SCC
		suodian(); 
		solve(); 
	}
	return 0;
}