时间限制：3 秒
内存限制：100 兆
特殊判题：否
提交：1362
解决：640
- 题目描述：
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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.
- 输入：
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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)
- 输出：
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The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.
总感觉这个Mr Wang是个恋童癖有木有==||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 }