Proving Equivalences Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6640 Accepted Submission(s): 2299
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:
A is invertible.Ax = b has exactly one solution for every n × 1 matrix b.Ax = b is consistent for every n × 1 matrix b.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
【分析】 强连通分量缩点 在缩点图上找 入度=0的个数p 和 出度=0的个数q，答案就是max(p,q)… 至于为什么…我的感觉是要从出度为0的点向入度为0的点连边就好了…
ps:整个图被缩为一个点，ans=0
【代码】
#include<iostream> #include<cstring> #include<cstdio> #include<stack> #define ll long long #define M(a) memset(a,0,sizeof a) #define fo(i,j,k) for(i=j;i<=k;i++) using namespace std; const int mxn=100005; stack <int> s; int n,m,cnt,tim,num,p,q; int dfn[mxn],low[mxn],head[mxn],bel[mxn],x[mxn],y[mxn]; bool vis[mxn],out[mxn],in[mxn]; struct edge {int to,next;} f[mxn<<1]; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();} while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar(); return x*f; } inline void add(int u,int v) { f[++cnt].to=v,f[cnt].next=head[u],head[u]=cnt; } inline void dfs(int u) { dfn[u]=low[u]=++tim; s.push(u),vis[u]=1; for(int i=head[u];i;i=f[i].next) { int v=f[i].to; if(!dfn[v]) dfs(v),low[u]=min(low[u],low[v]); else if(vis[v]) low[u]=min(low[u],dfn[v]); } if(dfn[u]==low[u]) { num++;int v; do { v=s.top(),s.pop(),vis[v]=0; bel[v]=num; }while(u!=v); } } int main() { int i,j,u,v,T; T=read(); while(T--) { cnt=tim=num=p=q=0; M(dfn),M(low),M(head),M(bel); M(out),M(in),M(vis); n=read(),m=read(); fo(i,1,m) { x[i]=read(),y[i]=read(); add(x[i],y[i]); } fo(i,1,n) if(!dfn[i]) dfs(i); fo(i,1,m) if(bel[x[i]]!=bel[y[i]]) out[bel[x[i]]]=1,in[bel[y[i]]]=1; if(num==1) {printf("0\n");continue;} fo(i,1,num) { if(!out[i]) p++; if(!in[i]) q++; } printf("%d\n",max(p,q)); } return 0; }