标签:
解题思路:将给出的男孩的关系合并后,另用一个数组a记录以find(i)为根节点的元素的个数,最后找出数组a的最大值
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others) Total Submission(s): 15861 Accepted Submission(s): 5843
#include<stdio.h>
#include<string.h>
int pre[10000010],a[10000010];
int find( int root)
{
if(root!=pre[root])
pre[root]=find(pre[root]);
return pre[root];
}
void unionroot( int root1,int root2)
{
int x,y;
x=find(root1);
y=find(root2);
if(x!=y)
pre[x]=y;
}
int main()
{
int n;
int root1,root2,x,y,k,i,max;
while(scanf("%d",&n)!=EOF)
{
max=-100000;
for(i=1;i<=10000010;i++)
a[i]=0;
for(i=1;i<=10000010;i++)
pre[i]=i;
while(n--)
{
scanf("%d %d",&root1,&root2);
x=find(root1);
y=find(root2);
unionroot(x,y);
}
for(i=1;i<=10000010;i++)
{
k=find(i);
a[k]++;//记录以find(i)为根节点的包含有多少 个元素
}
for(i=1;i<=10000010;i++)
{
if(a[i]>max)
max=a[i];
}
printf("%d\n",max);
}
}
标签:
原文地址:http://www.cnblogs.com/wuyuewoniu/p/4202541.html