HDOJ 题目3836 Equivalent Sets（强连通分量）

Equivalent Sets

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 104857/104857 K (Java/Others)
Total Submission(s): 3070    Accepted Submission(s): 1079

Problem Description

To prove two sets A and B are equivalent, we can first prove A is a subset of B, and then prove B is a subset of A, so finally we got that these two sets are equivalent.
You are to prove N sets are equivalent, using the method above: in each step you can prove a set X is a subset of another set Y, and there are also some sets that are already proven to be subsets of some other sets.
Now you want to know the minimum steps needed to get the problem proved.

 

Input

The input file contains multiple test cases, in each case, the first line contains two integers N <= 20000 and M <= 50000.
Next M lines, each line contains two integers X, Y, means set X in a subset of set Y.

 

Output

For each case, output a single integer: the minimum steps needed.

 

Sample Input

4 03 21 21 3

 

Sample Output

42
Hint
Case 2: First prove set 2 is a subset of set 1 and then prove set 3 is a subset of set 1.
 

 

Source

2011 Multi-University Training Contest 1 - Host by HNU

 

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 题目大意：问加几个边，让他成为一个强连通分量

ac代码

#include<stdio.h>#include<string.h>#include<queue>#include<string>#include<iostream>#define max(a,b) (a>b?a:b)#define min(a,b) (a<b?a:b)using namespace std;int inde[50020],outde[50020];int dfn[50020],low[50020],ins[50020],belong[50020],stack[50020],cnt,head[50050],top,taj,time;struct s{int u,v,next;}edge[50050];void add(int u,int v){edge[cnt].u=u;edge[cnt].v=v;edge[cnt].next=head[u];head[u]=cnt++;}void init(){cnt=0;memset(head,-1,sizeof(head));top=taj=0;memset(dfn,-1,sizeof(dfn));memset(belong,-1,sizeof(belong));memset(low,-1,sizeof(low));memset(ins,0,sizeof(ins));time=0;}void tarjan(int u){dfn[u]=low[u]=time++;ins[u]=1;stack[top++]=u;for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].v;if(dfn[v]==-1){tarjan(v);low[u]=min(low[u],low[v]);}elseif(ins[v]){low[u]=min(low[u],dfn[v]);}}if(dfn[u]==low[u]){int now;taj++;do{now=stack[--top];ins[now]=0;belong[now]=taj;}while(now!=u);}}int n,m;int main(){while(scanf("%d%d",&n,&m)!=EOF){int i;init();for(i=0;i<m;i++){int a,b;scanf("%d%d",&a,&b);add(a,b);}for(i=1;i<=n;i++)if(dfn[i]==-1)tarjan(i);if(n<1||taj==1){printf("0\n");continue;}int ans1,ans2;ans1=ans2=0;memset(inde,0,sizeof(inde));memset(outde,0,sizeof(outde));for(i=0;i<cnt;i++){int u=edge[i].u;int v=edge[i].v;if(belong[u]!=belong[v]){inde[belong[v]]++;outde[belong[u]]++;}}for(i=1;i<=taj;i++){if(inde[i]==0)ans1++;if(outde[i]==0)ans2++;}printf("%d\n",max(ans1,ans2));}}