HDU 2767 Proving Equivalences (強聯通)

標籤：圖論   強聯通   

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 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

 
SourceNWERC 2008 

題意：

要證明等價性（要求所有命題都是等價的），現已給出部分證明（u->v），問最少還需多少步才能完成目標。

分析：

我們的目標是（沒有蛀牙）使得整個圖是強聯通的，已經有部分有向邊u->v。我們先用強聯通縮點，得到一個有向非循環圖，設入度為0的點有a個，出度為0的點有b個，我們只要max(a,b)步就能完成目標（數學歸納法可證）。

#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 20007using 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/文檔/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;}