标签：beginning name clu exactly more import orm dir turn
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
4 1 2 3 4 5 6 1 6 4 1 2 3 4 5 6 7 8
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.
题意：王老师要找一些男生帮助他完成一项工程。要求最后挑选出的男生之间都是朋友关系，可以说直接的，也可以是间接地。问最多可以挑选出几个 男生（最少挑一个)。
代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 int num[1000000],par[1000000]; 6 void init() 7 { 8 for(int i=1;i<1000000;i++) 9 { par[i]=i; 10 num[i]=1; 11 } 12 } 13 int find(int x) 14 { 15 if(par[x]!=x) 16 return par[x]=find(par[x]); 17 return x; 18 19 } 20 void unite(int a,int b) 21 { 22 int fa=find(a); 23 int fb=find(b); 24 if(fa!=fb) 25 {par[fa]=fb; 26 num[fb]+=num[fa]; //这个地方注意，第一次num【fb】写成了b， 27 28 } 29 } 30 int main() 31 { 32 int a,b,n; 33 while(scanf("%d",&n)!=EOF) 34 {if(n==0) 35 {printf("1\n"); 36 continue; 37 38 } 39 init(); 40 int max=0; 41 while(n--) 42 { 43 scanf("%d %d",&a,&b); 44 if(a>max) max=a; 45 if(b>max) max=b; 46 unite(a,b); 47 } 48 int m=0; 49 for(int i=1;i<=max;i++) 50 {if(num[i]>m) 51 m=num[i]; 52 53 } 54 printf("%d\n",m); 55 56 57 58 } 59 return 0; 60 }
标签：beginning name clu exactly more import orm dir turn
原文地址：http://www.cnblogs.com/hss-521/p/7277109.html
