HUD——T 3836 Equivalent Sets

http://acm.hdu.edu.cn/showproblem.php?pid=3836

Time Limit: 12000/4000 MS (Java/Others) Memory Limit: 104857/104857 K (Java/Others)
Total Submission(s): 4802 Accepted Submission(s): 1725

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 0 3 2 1 2 1 3

Sample Output

4 2

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

Recommend

xubiao | We have carefully selected several similar problems for you: 3835 3828 3834 3830 3829

题意：求最少连几条边可以使整个图成为强连通图

可以先将图缩点，然后统计新图中入读==0，和出度==0 的点的个数，因为使加边最少，

所以应该是先给出度==0的点连一条向入读==0的点得边，然后再加上多余的（入读==0||出度==0）的点

ans=max(入读==0的点数,出读==0的点数)

 1 #include <algorithm>
 2 #include <cstring>
 3 #include <cstdio>
 4 
 5 using namespace std;
 6 
 7 const int N(20000+5);
 8 const int M(50000+5);
 9 int n,m;
10 
11 int head[N],sumedge;
12 struct Edge
13 {
14     int v,next;
15     Edge(int v=0,int next=0):v(v),next(next){} 
16 }edge[M];
17 inline void ins(int u,int v)
18 {
19     edge[++sumedge]=Edge(v,head[u]);
20     head[u]=sumedge;
21 }
22 
23 int tim,dfn[N],low[N];
24 int top,Stack[N],instack[N];
25 int sumcol,col[N];
26 void DFS(int now)
27 {
28     dfn[now]=low[now]=++tim;
29     Stack[++top]=now; instack[now]=1;
30     for(int i=head[now];i;i=edge[i].next)
31     {
32         int v=edge[i].v;
33         if(!dfn[v])  DFS(v),low[now]=min(low[now],low[v]);
34         else if(instack[v]) low[now]=min(low[now],dfn[v]);
35     }
36     if(dfn[now]==low[now])
37     {
38         col[now]=++sumcol;
39         for(;Stack[top]!=now;top--)
40         {
41             col[Stack[top]]=sumcol;
42             instack[Stack[top]]=0;
43         }
44         instack[now]=0; top--;
45     }
46 }
47 
48 int ans,ans1,ans2,rd[N],cd[N];
49 inline void init()
50 {
51     top=ans=ans1=ans2=tim=sumcol=sumedge=0;
52     memset(rd,0,sizeof(rd));
53     memset(cd,0,sizeof(cd));
54     memset(low,0,sizeof(low));
55     memset(dfn,0,sizeof(dfn));
56     memset(head,0,sizeof(head));
57     memset(Stack,0,sizeof(Stack));
58     memset(instack,0,sizeof(instack));
59 }
60 
61 int main()
62 {
63     for(;~scanf("%d%d",&n,&m);init())
64     {
65         for(int u,v,i=1;i<=m;i++)
66             scanf("%d%d",&u,&v),ins(u,v);
67         for(int i=1;i<=n;i++)
68             if(!dfn[i]) DFS(i);
69         for(int u=1;u<=n;u++)
70             for(int i=head[u];i;i=edge[i].next)
71             {
72                 int v=edge[i].v;
73                 if(col[u]==col[v]) continue;
74                 rd[col[v]]++; cd[col[u]]++;
75             }
76         for(int i=1;i<=sumcol;i++)
77         {
78             if(!cd[i]) ans1++;
79             if(!rd[i]) ans2++;
80         }
81         ans=max(ans1,ans2);
82         if(sumcol==1) ans=0;
83         printf("%d
",ans);
84     }
85     return 0;
86 }

——每当你想要放弃的时候，就想想是为了什么才一路坚持到现在。