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
MTK Android 如何获取系统权限
查看>>
MySQL - 4种基本索引、聚簇索引和非聚索引、索引失效情况、SQL 优化
查看>>
MySQL - ERROR 1406
查看>>
mysql - 视图
查看>>
MySQL - 解读MySQL事务与锁机制
查看>>
MTTR、MTBF、MTTF的大白话理解
查看>>
mt_rand
查看>>
mysql -存储过程
查看>>
mysql /*! 50100 ... */ 条件编译
查看>>
mudbox卸载/完美解决安装失败/如何彻底卸载清除干净mudbox各种残留注册表和文件的方法...
查看>>
mysql 1264_关于mysql 出现 1264 Out of range value for column 错误的解决办法
查看>>
mysql 1593_Linux高可用(HA)之MySQL主从复制中出现1593错误码的低级错误
查看>>
mysql 5.6 修改端口_mysql5.6.24怎么修改端口号
查看>>
MySQL 8.0 恢复孤立文件每表ibd文件
查看>>
MySQL 8.0开始Group by不再排序
查看>>
mysql ansi nulls_SET ANSI_NULLS ON SET QUOTED_IDENTIFIER ON 什么意思
查看>>
multi swiper bug solution
查看>>
MySQL Binlog 日志监听与 Spring 集成实战
查看>>
MySQL binlog三种模式
查看>>
multi-angle cosine and sines
查看>>