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事务隔离级别:读未提交、读已提交、可重复读和串行
查看>>
webpack css文件处理
查看>>
mysql二进制包安装和遇到的问题
查看>>
MySql二进制日志的应用及恢復
查看>>
mysql互换表中两列数据方法
查看>>
mysql五补充部分:SQL逻辑查询语句执行顺序
查看>>
mysql交互式连接&非交互式连接
查看>>
MySQL什么情况下会导致索引失效
查看>>
Mysql什么时候建索引
查看>>
MySql从入门到精通
查看>>
MYSQL从入门到精通(一)
查看>>
MYSQL从入门到精通(二)
查看>>
mysql以下日期函数正确的_mysql 日期函数
查看>>
mysql以服务方式运行
查看>>
mysql优化--索引原理
查看>>
MySQL优化之BTree索引使用规则
查看>>
MySQL优化之推荐使用规范
查看>>
Webpack Critical CSS 提取与内联教程
查看>>
mysql优化概述(范式.索引.定位慢查询)
查看>>
MySQL优化的一些需要注意的地方
查看>>