Proving Equivalences (hdu 2767 強聯通縮點)

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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 DescriptionConsider 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?

 
InputOn 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. 
OutputPer 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

24 03 21 21 3

 
Sample Output

42

 
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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題意：n個點m條邊，問最少加入多少條邊使得整個圖聯通。

思路：先Tarjan求強聯通分量，縮點，再求縮點後的點的入度和出度，入讀為0的點的個數為a。出度為0的點的個數為b，ans=max（a。b）

代碼：

#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;//點數const int MAXM = 500050;//邊數struct Edge{    int to,next;}edge[MAXM];int head[MAXN],tot;int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];//Belong數組的值是1~sccint Index,top;int scc;//強聯通分量的個數bool Instack[MAXN];int num[MAXN];//各個強聯通分量包括的點的個數。數組編號為1~scc//num數組不一定須要，結合實際情況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++)      //點的編號從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){   //假設強連通個數為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;}

Proving Equivalences (hdu 2767 強聯通縮點)