More is better(并查集注意节约时间，超时)

最新推荐文章于 2023-02-19 14:52:18 发布

原创最新推荐文章于 2023-02-19 14:52:18 发布 · 947 阅读

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本文介绍了一个计算最大好友群数量的问题，通过使用并查集算法来解决。具体地，文章给出了一段C语言代码示例，该代码实现了如何通过并查集来合并不同的好友群，并找出最大的好友群包含的人数。

点击打开链接

错：

1、超时，解决方案，

if(num[fx]>num[fy])
        {
            int t=fy;
            fy=fx;
            fx=t;
        }
        set[fx]=fy;

2、wrong answer 错在max赋值应该是1，错成-1，没有考虑所有点都不联通的情况

1、题目：

More is better
Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 7927    Accepted Submission(s): 2942


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
4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8
 

Sample Output
4
2

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.
  
 

Author
lxlcrystal@TJU
 

Source
HDU 2007 Programming Contest - Final 
 

Recommend
lcy

2、代码：

#include<stdio.h>
int set[10000005];
int num[10000005];
int ans[10000005];
int max;
int find(int x)
{
    int r=x;
    while(set[r]!=r)
    r=set[r];
    return r;
}
void merge(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
    {
        if(num[fx]>num[fy])
        {
            int t=fy;
            fy=fx;
            fx=t;
        }
        set[fx]=fy;
        num[fy]=num[fx]+num[fy];
        if(num[fy]>max)
        max=num[fy];
    }
}
int main()
{
    int n,a,b;
    while(scanf("%d",&n)!=EOF)
    {
        max=1;//错在max=-1，如果赋值为-1，且所有点都没联系，则merge函数中if永远进不去
        for(int i=1;i< 10000005;i++)
        {
            set[i]=i;
            num[i]=1;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&a,&b);
            merge(a,b);
        }
        printf("%d\n",max);
    }
    return 0;
}