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hdu 1856

2024-08-02 09:16:17   218人阅读

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 15434    Accepted Submission(s): 5667


Problem Description
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.
 

 

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
41 23 45 61 641 23 45 67 8
 

 

Sample Output
42
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.
 

 

Author
lxlcrystal@TJU
 

 

Source
HDU 2007 Programming Contest - Final
 

 

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#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<cstdlib>#include<algorithm>#define N 10000000int father[N],num[N];void initial()/*初始化*/{int i;for(i=1;i<=N;i++){father[i]=i;num[i]=1;/*开始时数量都为1，根节点为自己*/}}int find(int x) /*寻找根节点*/{if(father[x]!=x)father[x]=find(father[x]);return father[x];}void merge(int a,int b)/*合并a和b*/{int p=find(a);int q=find(b);if(p!=q){father[p]=q;num[q]+=num[p];/*合并集合中元素个数*/}}int main(){int n,a,b,i,sum,max;while(scanf("%d",&n)!=EOF){if(n==0){printf("1\n");continue;}max=0;initial(); /*初始化*/for(i=0;i<n;i++){scanf("%d%d",&a,&b);if(a>max)max=a;if(b>max)max=b;merge(a,b); /*合并集合*/}int Max=0;for(i=1;i<=max;i++)if(num[i]>Max) /*查找最大值*/Max=num[i];printf("%d\n",Max);}//system("pause");return 0;}

　　

hdu 1856


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