题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1856

More is better

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

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.
 

一定要细致看题目啊！

！！

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)

时间限制尽管在1s 可是内存放的非常宽啊！

。 所以什么数组存不下等等的顾虑都抛开吧，大胆的开数组！

【代码】

#include <cstdio>
const int maxn= 10000000+10;
int father[maxn];
int num[maxn];
int Find(int x){ //路径压缩迭代版本号
    int root = x;
    while(root!=father[root]) 
        root = father[root];//肇东根节点
    while(x!=root){ //将这棵树上的节点都指向根节点
        int tmp = father[x];
        father[x] = root;
        x= tmp;
    }
    return root;
}
void Union(int a,int b){
    int p=Find(a);
    int q=Find(b);
    if(p!=q){
        father[p]=q;
        num[q]+=num[p]; //将合并后的数量加到 父节点上
    }
}
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        if(n==0){
            puts("1");continue;
        }
        for(int i=1;i<=maxn;i++){
            num[i]=1;
            father[i]=i;
        }
        int a,b;
        int Max=0;
        for(int i=0;i<n;i++){
            scanf("%d%d",&a,&b);
            if(a>Max)
                Max=a;
            if(b>Max)
                Max=b;
            Union(a,b);
        }
        //printf("sd");
        int max=0;
         for(int i=1;i<=Max;i++){
             if(num[i]>max)
                 max=num[i];
            // printf("bug\n");
         }
        printf("%d\n",max);
    }
    return 0;
}