hdu 2767 Proving Equivalences（至少加几条边让整个图变成强连通）

最新推荐文章于 2021-10-14 21:04:13 发布

原创最新推荐文章于 2021-10-14 21:04:13 发布 · 793 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#tarjan

hdu 同时被 2 个专栏收录

170 篇文章

订阅专栏

强连通双连通 LCA

24 篇文章

订阅专栏

本文解析了一道关于证明矩阵等价性的线性代数题目，通过构建图模型并使用Tarjan算法来解决最少边添加问题，实现证明所有声明等价的目标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Proving Equivalences

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2088 Accepted Submission(s): 796

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

Sample Output

4
2
 
题意：同hdu 3836是一样的， 问至少加几条边让整个图变成强连通。
思路：这里用tarjan算法缩点。
 
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdlib>
#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)

using namespace std;

const int maxn=20005;
const int maxm=50005;

struct node
{
    int v,next;
} edge[maxm];
int head[maxn],low[maxn],dfn[maxn];
int stack[maxn],scc[maxn],in[maxn],out[maxn];
bool ins[maxn];
int n,m,top,cnt,snum,num;
void init()
{
    memset(head,-1,sizeof(head));
    num=0;
}
void add(int u,int v)
{
    edge[num].v=v;
    edge[num].next=head[u];
    head[u]=num++;
}
void input()
{
    int a,b;
    scanf("%d%d",&n,&m);
    while(m--)
    {
        scanf("%d%d",&a,&b);
        add(a,b);
    }
}
void dfs(int u)
{
    int x;
    dfn[u]=low[u]=++cnt;
    stack[top++]=u;
    ins[u]=true;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(!dfn[v])
        {
            dfs(v);
            low[u]=min(low[u],low[v]);
        }
        else if(ins[v]) low[u]=min(low[u],dfn[v]);
    }
    if(low[u]==dfn[u])
    {
        ++snum;
        do{
            x=stack[--top];
            ins[x]=false;
            scc[x]=snum;
        }while(x!=u);
    }
}
void tarjan()
{
    memset(dfn,0,sizeof(dfn));
    memset(ins,false,sizeof(ins));
    snum=cnt=top=0;
    for(int i=1;i<=n;i++)
    if(!dfn[i]) dfs(i);
}
void solve()
{
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    for(int u=1;u<=n;u++)
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(scc[u]!=scc[v])
        {
            out[scc[u]]++;
            in[scc[v]]++;
        }
    }
    if(snum==1) { printf("0\n");return;}
    int ans1=0,ans2=0;
    for(int i=1;i<=snum;i++)
    {
    if(!in[i]) ans1++;
    if(!out[i]) ans2++;
    }
    printf("%d\n",max(ans1,ans2));
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        input();
        tarjan();
        solve();
    }
    return 0;
}