Proving Equivalences
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7346 Accepted Submission(s): 2539
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?
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.
* 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.
* 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
找出所有的强连通分量， 然后缩成一个点，然后统计缩点之后的新图的出度为0的点的个数（记为cntOut),和入度为0的点的个数（记为cntIn)
那么要加边的条数就是max(cntOut,cntIn)
这个为什么呢？？ 因为，如果一个点的入度为0，那么说明这个点是不可达的，如果一个点的出度为0，那么说明这个点到其它点是不可达的。
为了解决这个情况，那么只要在出度为0的点（设为u）和入度为0的点之间连一条u-->v的边，那么就解决了这种情况。
不断的连边，只要一个点问题没解决就要连边， 所以是在两者之间取max 此段论述http://www.cnblogs.com/justPassBy/p/4678192.html转自这里
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N=20008; const int M=50008; int head[N],bl[N],q[N],dfn[N],low[N]; int tot,scnt,cnt,l,n,m; bool instack[N],ru[N],out[N]; struct node{ int to,next; }e[M]; void init(){ for(int i=0;i<=n;++i) { head[i]=-1; dfn[i]=instack[i]=0; ru[i]=out[i]=0; } l=tot=scnt=cnt=0; } void add(int u,int v){ e[tot].to=v; e[tot].next=head[u]; head[u]=tot++; } void Tajan(int u){ dfn[u]=low[u]=++cnt; instack[u]=1; q[l++]=u; for(int i=head[u];i+1;i=e[i].next){ int v=e[i].to; if(!dfn[v]) { Tajan(v); low[u]=min(low[u],low[v]); } else if(instack[v]&&dfn[v]<low[u]) low[u]=dfn[v]; } if(low[u]==dfn[u]){ int t; ++scnt; do{ t=q[--l]; instack[t]=0; bl[t]=scnt; }while(t!=u); } } int main(){ int u,v,T; for(scanf("%d",&T);T--;){ scanf("%d%d",&n,&m);init(); for(int i=1;i<=m;++i) { scanf("%d%d",&u,&v); add(u,v); } for(int i=1;i<=n;++i) if(!dfn[i]) Tajan(i); if(scnt==1) {puts("0");continue;} for(int i=1;i<=n;++i){ for(int j=head[i];j+1;j=e[j].next){ int v=e[j].to; if(bl[i]==bl[v]) continue; else { ru[bl[v]]=1; out[bl[i]]=1; } } } int ans1=0,ans2=0; for(int i=1;i<=scnt;++i){ if(!out[i]) ++ans1; if(!ru[i]) ++ans2; } printf("%d ",max(ans1,ans2)); } }