【强连通分量】Proving Equivalences

【题目链接】hdu-2767

【题目描述】

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?

【输入格式】

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.

【输出格式】

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.

【分析】

就是要求最少加多少条边可以使这个图变成强连通的。做完tarjan后，我们就找入度和出度为0的点，取最大值。

【代码】

#include <cstdio>

#include <iostream>

#include <cstring>

#include <algorithm>

#include <queue>

#include <cctype>

#include <cmath>

#include <time.h>

using namespace std;

const int maxm=50010;

const int maxn=20010;

struct Edge{

    int to,next;

}edge[maxm<<1];

int nedge,sum,dep,top,n,m,cnt;

int head[maxn],dfn[maxn],stack[maxm],low[maxm],od[maxm],vis[maxm],id[maxm];

int belong[maxn];

inline int read()

{

    int X=0,w=0; char ch=0;

    while(!isdigit(ch)) {w|=ch=='-';ch=getchar();}

    while(isdigit(ch)) X=(X<<3)+(X<<1)+(ch^48),ch=getchar();

    return w?-X:X;

}

void tarjan(int u)

{

    dfn[u]=low[u]=++dep;

    stack[top++]=u;

    vis[u]=1;

    for (int i=head[u];i!=-1;i=edge[i].next)

    {

        int v=edge[i].to;

        if (!dfn[v])

        {

            tarjan(v);

            low[u]=min(low[u],low[v]);

        }

        else

        {

            if (vis[v]) low[u]=min(low[u],dfn[v]);

        }

    }

    int j;

    if (dfn[u]==low[u])

    {

        sum++;

        do{

            j=stack[--top];

            belong[j]=sum;

            vis[j]=0;

        }while (u!=j);

    }

}

void add_edge(int a,int b)

{

    edge[nedge]=(Edge){b,head[a]}; head[a]=nedge++;

}

int main()

{

    int cas=read();

    while (cas--)

    {

        nedge=0;

        memset(head,-1,sizeof(head));

        memset(dfn,0,sizeof(dfn));

        memset(od,0,sizeof(od));

        memset(id,0,sizeof(id));

        memset(vis,0,sizeof(vis));

        memset(belong,0,sizeof(belong));

        sum=0,dep=0,top=0,cnt=0;

        n=read(),m=read();

        for (int i=1;i<=m;i++)

        {

            int a=read(),b=read();

            add_edge(a,b);

        }

        for (int i=1;i<=n;i++)

        {

            if (!dfn[i]) tarjan(i);

        }

        if (sum==1)

        {

            printf("0\n");

        }

        else

        {

            for (int i=1;i<=n;i++)

            {

                for (int j=head[i];j!=-1;j=edge[j].next)

                {

                    int v=edge[j].to;

                    if (belong[i]!=belong[v]) od[belong[i]]++,id[belong[v]]++;

                }

            }

            int idnum=0,odnum=0;

            for (int i=1;i<=sum;i++)

            {

                if (id[i]==0) idnum++;

                if (od[i]==0) odnum++;

            }

            int ans=max(idnum,odnum);

            printf("%d\n",ans);

        }

    }

    return 0;

}

巴特西

【强连通分量】Proving Equivalences

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