?

時間限制：3 秒

內存限制：100 兆

特殊判題：否

提交：1362

解決：640

題目描述：

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.

輸入：

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)

輸出：?

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


總感覺這個Mr Wang是個戀童癖有木有==||

?

  1 #include <iostream>
  2 
  3 using namespace std;
  4 
  5  
  6 
  7  
  8 
  9  
 10 
 11 struct LNode
 12 
 13 {
 14 
 15    LNode * lchild,*rchild;
 16 
 17    int data;
 18 
 19 };
 20 
 21  
 22 
 23  
 24 
 25 int Tree[10000001];
 26 
 27 int sum[10000001];
 28 
 29  
 30 
 31 int getroot(int x)
 32 
 33 {
 34 
 35     if(Tree[x]==-1) return x;
 36 
 37 else return getroot(Tree[x]);
 38 
 39 }
 40 
 41  
 42 
 43  
 44 
 45  
 46 
 47 int main()
 48 
 49 {
 50 
 51        int n;
 52 
 53    while(cin>>n)
 54 
 55    {
 56 
 57      int i;
 58 
 59  for(i=0;i<10000001;i++)
 60 
 61  {
 62 
 63     Tree[i]=-1;
 64 
 65 sum[i]=1;
 66 
 67  }
 68 
 69  
 70 
 71  
 72 
 73  
 74 
 75  for(i=0;i<n;i++)
 76 
 77  {
 78 
 79     int a,b;
 80 
 81 cin>>a>>b;
 82 
 83 a=getroot(a);
 84 
 85 b=getroot(b);
 86 
 87 if(a!=b)
 88 
 89 {
 90 
 91    Tree[a]=b;
 92 
 93    sum[b]+=sum[a];
 94 
 95 }
 96 
 97  }
 98 
 99  
100 
101  int temp=1;
102 
103  for(i=0;i<10000001;i++)
104 
105  {
106 
107    if(sum[i]>temp) temp=sum[i];
108 
109  }
110 
111  
112 
113  cout<<temp<<endl;
114 
115  
116 
117    }
118 
119         return 0;
120 
121 }

?

?

轉載于:https://www.cnblogs.com/xiaoyesoso/p/4265124.html

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