More is better（并查集）

最新推荐文章于 2021-10-25 19:11:05 发布

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#ACM #数据结构 #算法

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本文介绍了一个算法问题，即如何计算在特定条件下可能保留的最大人数。通过构建并查集解决复杂的人际关系网络问题，实现快速查找和合并操作，以确定能够保持相互为直接或间接朋友关系的最大群体规模。

Link：点击打开链接

Problem：

More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 15106 Accepted Submission(s): 5566

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

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

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code：

#include<stdio.h>
int pre[10000111],rank[10000111];
int n,ret;
void init()
{
int i;
for(i=1;i<=10000000;i++)
{
pre[i]=i;
rank[i]=1;
}
ret=0;
}
int find(int root)
{
int son,tmp;
son=root;
while(root!=pre[root])
{
root=pre[root];
}
while(son!=root)
{
tmp=pre[son];
pre[son]=root;
son=tmp;
}
return root;
}
void join(int root1,int root2)
{
int x,y;
x=find(root1);
y=find(root2);
if(x!=y)
{
if(rank[x]>rank[y])
{
pre[y]=x;
rank[x]+=rank[y];
if(ret<rank[x])
ret=rank[x];
}
else
{
pre[x]=y;
rank[y]+=rank[x];
if(ret<rank[y])
ret=rank[y];
}
}
}
int main()
{
int a,b;
while(scanf("%d",&n)!=EOF)
{
if(n==0)
{
printf("1\n");
continue;
}
init();
while(n--)
{
scanf("%d%d",&a,&b);
join(a,b);
}
printf("%d\n",ret);
}
return 0;
}

总结：注意n=0的情况！！！