Proving Equivalences (hdu 2767 强联通缩点)

Proving Equivalences

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

Total Submission(s): 3743    Accepted Submission(s): 1374

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

field=problem&key=NWERC+2008&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">NWERC 2008

 

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题意：n个点m条边，问最少加入多少条边使得整个图联通。

思路：先Tarjan求强联通分量，缩点，再求缩点后的点的入度和出度，入读为0的点的个数为a。出度为0的点的个数为b，ans=max（a。b）

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

const int MAXN = 20050;//点数
const int MAXM = 500050;//边数

struct Edge
{
    int to,next;
}edge[MAXM];

int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];//Belong数组的值是1~scc
int Index,top;
int scc;//强联通分量的个数
bool Instack[MAXN];
int num[MAXN];//各个强联通分量包括的点的个数。数组编号为1~scc
//num数组不一定须要，结合实际情况

void addedge(int u,int v)
{
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void Tarjan(int u)
{
    int v;
    Low[u]=DFN[u]=++Index;
    Stack[top++]=u;
    Instack[u]=true;
    for (int i=head[u];i+1;i=edge[i].next)
    {
        v=edge[i].to;
        if (!DFN[v])
        {
            Tarjan(v);
            if (Low[u]>Low[v]) Low[u]=Low[v];
        }
        else if (Instack[v]&&Low[u]>DFN[v])
            Low[u]=DFN[v];
    }
    if (Low[u]==DFN[u])
    {
        scc++;
        do{
            v=Stack[--top];
            Instack[v]=false;
            Belong[v]=scc;
            num[scc]++;
        }while (v!=u);
    }
}

void solve(int N)
{
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    memset(num,0,sizeof(num));
    Index=scc=top=0;
    for (int i=1;i<=N;i++)      //点的编号从1開始
        if (!DFN[i])
            Tarjan(i);
}

void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
}

int n,m;
int in[MAXN],out[MAXN];

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
#endif
    int i,j,u,v,t;
    sf(t);
    while (t--)
    {
        sff(n,m);
        if(n==1){   //特判1（n==1,m==0)
            printf("0\n");
            continue;
        }
        if(m==0){   //特判2( n==?

,m==0)
            printf("%d\n",n);
            continue;
        }
        init();
        for (i=0;i<m;i++)
        {
            sff(u,v);
            addedge(u,v);
        }
        solve(n);
        if(scc==1){   //假设强连通个数为1
            printf("0\n");
            continue;
        }
        mem(in,0);
        mem(out,0);
        for (int u=1;u<=n;u++)
        {
            for (i=head[u];i+1;i=edge[i].next)
            {
                int v=edge[i].to;
                if (Belong[u]!=Belong[v])
                {
                    out[Belong[u]]++;
                    in[Belong[v]]++;
                }
            }
        }
        int ans,a=0,b=0;
        for (i=1;i<=scc;i++)
        {
            if (out[i]==0)
                a++;
            if (in[i]==0)
                b++;
        }
        ans=max(a,b);
        pf("%d\n",ans);
    }
    return 0;
}