More is better--hdu1856（并查集）

最新推荐文章于 2021-10-25 19:11:05 发布

原创最新推荐文章于 2021-10-25 19:11:05 发布 · 336 阅读

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本文探讨了如何使用并查集算法解决在给定好友关系的情况下，找到能够保持最大数量朋友团的方法。通过解析输入的好友对，算法能够有效识别并合并朋友关系，最终确定最大的朋友团体。实例演示了算法应用过程，并提供了源代码实现。

More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 18707 Accepted Submission(s): 6873

Problem Description

Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements.

Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

Input

The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)

Output

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

Sample Input

1 2

3 4

5 6

1 6

1 2

3 4

5 6

7 8

Sample Output

Hint

A and B are friends(direct or indirect), B and C are friends(direct or indirect), then A and C are also friends(indirect). In the first sample {1,2,5,6} is the result. In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.

这个题的意思就是输入是好友的两人，两好友的盆友具有传递性！最后输出互相是好友最多的一群有多少人？

用并查集，具体来看注释部分！

 1 #include <iostream>
 2 #include<stdio.h>
 3 #include<string.h>
 4 #include<algorithm>
 5 #define maxn 100000
 6 using namespace std;
 7 int ran[maxn],per[maxn],via[maxn],num[maxn],n,sum;
 8 void init()
 9 {
10     int i;
11     for(i=1;i<=maxn;i++)
12     {
13         per[i]=i;
14         num[i] = 1;//记录每个根节点所包含的子节点
15     }
16 }
17 int find(int x)//查找函数
18 {
19     int t=x;
20     while(t!=per[t])
21      t=per[t];
22      int i=x,j;
23      while(i!=t)
24      {
25          j=per[i];
26          per[i]=t;
27          i=j;//压缩路径
28      }
29      return t;
30 }
31 void join(int x,int y)//合并函数，合并图
32 {
33     int fx=find(x);
34     int fy=find(y);
35     if(fx!=fy)
36     {
37         per[fx] = fy;
38         num[fy] += num[fx];
39         sum = max(sum, num[fy]);//求出各个树中子节点最多的
40     }
41     else
42         return ;
43 }
44 int main ()
45 {
46     int a,b,i,w,n;
47     while(scanf("%d",&n)!=EOF)
48     {
49         if(n == 0){
50             printf("1\n");//特殊坑爹数据
51             continue;
52         }
53         init();
54         w=n;
55         sum = 0;
56         while(w--)
57         {
58             scanf("%d%d",&a,&b);
59             via[a]=1;
60             via[b]=1;
61             join(a,b);
62         }
63     printf("%d\n", sum);
64     }
65     return 0;
66 }