pid=2767">http://acm.hdu.edu.cn/showproblem.php?pid=2767

Proving Equivalences

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

Total Submission(s): 2926    Accepted Submission(s): 1100

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
 
Source
 

题意:

要证明等价性(要求全部命题都是等价的),现已给出部分证明(u->v),问最少还需多少步才干完毕目标。

分析:

我们的目标是(没有蛀牙)使得整个图是强联通的,已经有部分有向边u->v。我们先用强联通缩点,得到一个有向无环图,设入度为0的点有a个,出度为0的点有b个。我们仅仅要max(a,b)步就能完毕目标(数学归纳法可证)。

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<ctime>
#include<cctype>
#include<cmath>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<list>
#include<vector>
#include<map>
#include<set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#define maxm 50000
#define maxn 20007 using namespace std; int in[maxn],out[maxn];
int fir[maxn];
int u[maxm],v[maxm],nex[maxm];
int sccno[maxn],pre[maxn],low[maxn];
int st[maxn],top;
int scc_cnt,dfs_clock;
int n,m; void tarjan_dfs(int _u)
{
pre[_u]=low[_u]=++dfs_clock;
st[++top]=_u;
for (int e=fir[_u];~e;e=nex[e])
{
int _v=v[e];
if (!pre[_v])
{
tarjan_dfs(_v);
low[_u]=min(low[_u],low[_v]);
}
else
{
if (!sccno[_v])
{
low[_u]=min(low[_u],pre[_v]);
}
}
} if (pre[_u]==low[_u])
{
++scc_cnt;
while (true)
{
int x=st[top--];
sccno[x]=scc_cnt;
if (x==_u) break;
}
}
} void find_scc()
{
scc_cnt=dfs_clock=0;top=-1;
memset(pre,0,sizeof pre);
memset(sccno,0,sizeof sccno);
for (int i=1;i<=n;i++)
{
if (!pre[i]) tarjan_dfs(i);
}
} int main()
{
#ifndef ONLINE_JUDGE
freopen("/home/fcbruce/文档/code/t","r",stdin);
#endif // ONLINE_JUDGE int T_T;
scanf("%d",&T_T);
while (T_T--)
{ scanf("%d %d",&n,&m);
memset(fir,-1,sizeof fir);
for (itn i=0;i<m;i++)
{
scanf("%d %d",&u[i],&v[i]);
nex[i]=fir[u[i]];
fir[u[i]]=i;
} find_scc(); if (scc_cnt==1)
{
printf("%d\n",0);
continue;
} memset(in,0,sizeof in);
memset(out,0,sizeof out);
for (int i=0;i<m;i++)
{
if (sccno[u[i]]==sccno[v[i]]) continue; in[sccno[v[i]]]++;
out[sccno[u[i]]]++;
} int a=0,b=0;
for (itn i=1;i<=scc_cnt;i++)
{
if (in[i]==0) a++;
if (out[i]==0) b++;
} printf("%d\n",max(a,b));
} return 0;
}

HDU 2767 Proving Equivalences (强联通)的更多相关文章

  1. HDU 2767 Proving Equivalences(强连通 Tarjan+缩点)

    Consider the following exercise, found in a generic linear algebra textbook. Let A be an n × n matri ...

  2. hdu 2767 Proving Equivalences

    Proving Equivalences 题意:输入一个有向图(强连通图就是定义在有向图上的),有n(1 ≤ n ≤ 20000)个节点和m(0 ≤ m ≤ 50000)条有向边:问添加几条边可使图变 ...

  3. HDU 2767 Proving Equivalences(至少增加多少条边使得有向图变成强连通图)

    Proving Equivalences Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Oth ...

  4. HDU 2767 Proving Equivalences (Tarjan)

    Proving Equivalences Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other ...

  5. hdu 2767 Proving Equivalences(tarjan缩点)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2767 题意:问最少加多少边可以让所有点都相互连通. 题解:如果强连通分量就1个直接输出0,否者输出入度 ...

  6. hdu 2767 Proving Equivalences 强连通缩点

    给出n个命题,m个推导,问最少添加多少条推导,能够使全部命题都能等价(两两都能互推) 既给出有向图,最少加多少边,使得原图变成强连通. 首先强连通缩点,对于新图,每一个点都至少要有一条出去的边和一条进 ...

  7. HDU 2767:Proving Equivalences(强连通)

    题意: 一个有向图,问最少加几条边,能让它强连通 方法: 1:tarjan 缩点 2:采用如下构造法: 缩点后的图找到所有头结点和尾结点,那么,可以这么构造:把所有的尾结点连一条边到头结点,就必然可以 ...

  8. HDU 2767-Proving Equivalences(强联通+缩点)

    题目地址:pid=2767">HDU 2767 题意:给一张有向图.求最少加几条边使这个图强连通. 思路:先求这张图的强连通分量.假设为1.则输出0(证明该图不须要加边已经是强连通的了 ...

  9. hdoj 2767 Proving Equivalences【求scc&&缩点】【求最少添加多少条边使这个图成为一个scc】

    Proving Equivalences Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Oth ...

随机推荐

  1. Guns V2.5

    Guns V2.5 新版Guns基于SpringBoot全面升级,完美整合springmvc + shiro + MyBatis 通用 Mapper + 分页插件 PageHelper + beetl ...

  2. js 数组遍历

    map.filter.forEach.every.some http://www.runoob.com/jsref/jsref-obj-array.html 1.在字符串中使用 map 在一个Stri ...

  3. Mysql建立触发器

    DELIMITER $$ CREATE /*!50017 DEFINER = 'root'@'%' */ TRIGGER `AddTransferAccountLog` AFTER INSERT ON ...

  4. 百度之星初赛(A)——T6

    度度熊的01世界 Problem Description 度度熊是一个喜欢计算机的孩子,在计算机的世界中,所有事物实际上都只由0和1组成. 现在给你一个n*m的图像,你需要分辨他究竟是0,还是1,或者 ...

  5. EF4学习链接

    原文发布时间为:2011-09-23 -- 来源于本人的百度文章 [由搬家工具导入] 1.Fluent API 的方式定义与数据库映射 2.利用特性实现与数据库的映射 3.EF的一些公约的介绍 4.E ...

  6. Python数据结构——栈

    栈是一种特殊的列表,栈内的元素只能通过列表的一端访问,这一端称为栈顶.栈被称为一种后入先出(LIFO,last-in-first-out)的数据结构. 由于栈具有后入先出的特点,所以任何不在栈顶的元素 ...

  7. 判断dataset是否被修改—DataSet.HasChanges 方法

    DataSet.HasChanges 方法 获取一个值,该值指示 DataSet 是否有更改,包括新增行.已删除的行或已修改的行. 命名空间:   System.Data程序集:  System.Da ...

  8. RobotFramework自动化4-批量操作案例【转载】

    本篇转自博客:上海-悠悠 原文地址:http://www.cnblogs.com/yoyoketang/tag/robotframework/ 前言 有时候一个页面上有多个对象需要操作,如果一个个去定 ...

  9. Table is marked as crashed and should be repaire (

    https://www.cnblogs.com/cxchanpin/p/6894747.html

  10. UVA 562 Dividing coins【01背包 / 有一堆各种面值的硬币,将所有硬币分成两堆,使得两堆的总值之差尽可能小】

    It's commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nic ...