随笔- 89 文章- 22 评论- 74 阅读- 19万

hdu Disjoint-set problems

Disjoint-set namely Union-Find set!

hdu 3172 virtual friends http://acm.hdu.edu.cn/showproblem.php?pid=3172

Yes! A very interesting problem!

Whenever a friendship is formed,printf the number of the peoplein the social network.

OK, I failed lots of time. And all are "Time Limit Exceeded".

But to my surprise! I just replace the "cin" with “scanf” ,"cout" with "printf" in the code,and accepted!

learned: scanf/printf is more efficient than cin/cout;

View Code

 1 #include<iostream>
 2 #include<cstring>
 3 #include<string>
 4 #include<map>
 5 #include<cstdio>
 6 using namespace std;
 7 int f[200005],num[200005];
 8 void sequence()
 9 {
10     int i;
11     for(i=1;i<200005;i++)
12     {
13         f[i]=i;
14         num[i]=1;
15     }
16 }
17 int find(int x)
18 {
19     if(x!=f[x])
20         f[x]=find(f[x]);
21     return f[x];
22 
23 }
24 void Union(int x,int y)
25 {
26      x=find(x);
27      y=find(y);
28      if(x==y)
29      {
30          printf("%d\n",num[x]);
31      }
32      else
33      {
34          f[y]=x;
35          num[x]+=num[y];
36         printf("%d\n",num[x]);
37      }
38 }
39 int main()
40 {
41     int t;
42     map<string,int>match;
43     while(scanf("%d",&t)!=EOF)
44     {
45         int k;
46         string name1,name2;
47         while(t--)
48         {
49             sequence();
50             match.clear();
51             scanf("%d",&k);
52             int s=1;
53             while(k--)
54             {
55                 cin>>name1>>name2;
56                 if(match.find(name1)==match.end())
57                 {
58                     s++;match[name1]=s;
59                 }
60                 if(match.find(name2)==match.end())
61                 {
62                     s++;match[name2]=s;
63                 }
64                 Union(match[name1],match[name2]);
65             }
66         }
67     }
68     return 0;
69 }
70 
71             
72             
73 
74