博客
关于我
hdu2767(强连通分量+缩点)
阅读量:245 次
发布时间:2019-03-01

本文共 3084 字,大约阅读时间需要 10 分钟。

Proving Equivalences

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7367 Accepted Submission(s): 2547

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
对整个图求一次强连通分量,如果强连通分量为1则直接输出0,否则进行缩点(啥叫缩点:我们求强连通分量时,给每个顶点做一个标记,标记该顶点属于哪个强联通分量,然后属于同一个强连通分量的点就可以看作同一个点了。这就是所谓的“缩点”)对整个图缩点后这个图就变成了有向无环图,假设这个有向无环图入度为零的点有a个,出度为零的点有b个,这结果为max(a,b)(这个结论可以画个图推一推)

#include 
using namespace std;const int N=20010;vector
> vec(N);int low[N],dfn[N],Stack[N],belong[N];bool InStack[N];int in[N],out[N];int Index,top,ans;void Tarjan(int u){ low[u]=dfn[u]=(++Index); Stack[top++]=u; InStack[u]=true; for(int i=0;i
low[v]){ low[u]=low[v]; } } else if(InStack[v]&&low[u]>dfn[v]){ low[u]=dfn[v]; } } if(low[u]==dfn[u]){ int v; ans++; do{ v=Stack[--top]; belong[v]=ans; InStack[v]=false; } while(v!=u); }}void init(int n){ for(int i=1;i<=n;i++){ vec[i].clear(); } memset(InStack,false,sizeof(InStack)); memset(belong,0,sizeof(belong)); memset(dfn,0,sizeof(dfn)); memset(in,0,sizeof(in)); memset(out,0,sizeof(out)); Index=top=ans=0;}int main(){ int T; scanf("%d",&T); while(T--){ int n,m; scanf("%d %d",&n,&m); init(n); for(int i=0;i

转载地址:http://nnfx.baihongyu.com/

你可能感兴趣的文章
MySQL 设置数据库的隔离级别
查看>>
MySQL 证明为什么用limit时,offset很大会影响性能
查看>>
Mysql 语句操作索引SQL语句
查看>>
MySQL 误操作后数据恢复(update,delete忘加where条件)
查看>>
MySQL 调优/优化的 101 个建议!
查看>>
mysql 转义字符用法_MySql 转义字符的使用说明
查看>>
mysql 输入密码秒退
查看>>
mysql 递归查找父节点_MySQL递归查询树状表的子节点、父节点具体实现
查看>>
mysql 通过查看mysql 配置参数、状态来优化你的mysql
查看>>
mysql 里对root及普通用户赋权及更改密码的一些命令
查看>>
Mysql 重置自增列的开始序号
查看>>
mysql 锁机制 mvcc_Mysql性能优化-事务、锁和MVCC
查看>>
MySQL 错误
查看>>
mysql 随机数 rand使用
查看>>
MySQL 面试题汇总
查看>>
MySQL 面试,必须掌握的 8 大核心点
查看>>
MySQL 高可用性之keepalived+mysql双主
查看>>
mysql 默认事务隔离级别下锁分析
查看>>
Mysql--逻辑架构
查看>>
MySql-2019-4-21-复习
查看>>