问题 D: More is better

时间限制: 1 Sec  内存限制: 128 MB
提交: 546  解决: 203
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题目描述

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.

输入

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)

输出

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

样例输入

3
1 3
1 5
2 5
4
3 2
3 4
1 6
2 6

样例输出

4
5

 

#include
#include
#include
using namespace std;
const int n=10000001;
//数组的下标是可以开到10e7次方级别的。 
int father[n];
int isRoot[n];

int findFather(int x){
	
	if(x==father[x])
	 return x;
	
	int f=findFather(father[x]);
	father[x]=f;
	return f;
	
}
void Union(int a,int b){
	
	int fA=findFather(a);
	int fB=findFather(b);
	
	if(fA!=fB)
	 father[fA]=fB;
	
}

int main(){
	
  	int m,a,b;
  	while(cin>>m){
  		
  		memset(isRoot,0,sizeof(isRoot));
		for(int i=1;i>a>>b;
			Union(a,b);			
		}  		
  		
  		for(int i=1;i

同样的思路换成vector就超时了。。。。。 

#include
#include
#include
#include
using namespace std;
const int n=10000001;

vector father(n);
vector isRoot(n,0);

int findFather(int x){
	
	if(x==father[x])
	 return x;
	
	int f=findFather(father[x]);
	father[x]=f;
	return f;
	
}
void Union(int a,int b){
	
	int fA=findFather(a);
	int fB=findFather(b);
	
	if(fA!=fB)
	 father[fA]=fB;
	
}

int main(){
	
  	int m,a,b;
  	while(cin>>m){
  			 
		for(int i=1;i>a>>b;
			Union(a,b);			
		}  		
  		
  		for(int i=1;i