More is better 查并集并且找最大连通分量

左手的ㄟ右手 2024-02-18 22:54 37阅读 0赞

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 <td> <h2>More is better</h2> Time Limit: 5000/1000 MS (Java/Others)&nbsp;&nbsp;&nbsp;&nbsp;Memory Limit: 327680/102400 K (Java/Others) Total Submission(s): 34171&nbsp;&nbsp;&nbsp;&nbsp;Accepted Submission(s): 12094 &nbsp; 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. &nbsp; &nbsp; 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) &nbsp; &nbsp; Output The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep. &nbsp; &nbsp; Sample Input <pre>&nbsp;</pre> 4 1 2 3 4 5 6 1 6 4 1 2 3 4 5 6 7 8 &nbsp; &nbsp; Sample Output <pre>&nbsp;</pre> 4 2 Hint </td> 
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\#include<iostream> 
\#include<string.h> 
\#include<algorithm> 
using namespace std; 
int a\[10000001\],b\[100001\]; 
struct kk 
\{ 
int l,r; 
\}s\[100001\]; 
int cmp(int x,int y) 
\{ 
return x>y; 
\} 
int find(int x) 
\{ 
int r=x,j; 
while(x!=a\[x\]) 
x=a\[x\]; 
while(r!=x) 
\{ 
j=a\[r\]; 
a\[r\]=x; 
r=j; 
\} 
return x; 
\} 
int join(int x,int y) 
\{ 
int fx=find(x); 
int fy=find(y); 
if(fx!=fy) 
\{ 
a\[fy\]=a\[fx\]; 从今天开始，fy就是我a\[fx\]派的啦 
b\[a\[fx\]\]+=b\[fy\]; 你派的人马也归我了 
return 1; 
\} 
else 
return 0; 
\} 
int main() 
\{ 
int n,x,y,i,j,t; 
while(~scanf("%d",&n)) 
\{ 
int m=0; 
if(n==0) 
\{ 
cout<<1<<endl; 
continue; 
\} 
t=0; 
int max=0; 
memset(b,0,sizeof(b)); 
for(i=1;i<=n;i++) 
\{ 
cin>>s\[i\].l>>s\[i\].r; 
if(s\[i\].l>max) 
max=s\[i\].l; 
if(s\[i\].r>max) 
max=s\[i\].r; 
\} 
for(i=1;i<=max;i++) 
\{ 
a\[i\]=i; 
b\[i\]=1; 原始武林 ，各成一派 
\} 
 
for(i=1;i<=n;i++) 
\{ 
join(s\[i\].l,s\[i\].r); 
\} 
for(i=1;i<=max;i++) 
\{ 
if(b\[i\]>m) 找最大帮派 
\{ 
m=b\[i\]; 
\} 
\} 
cout<<m<<endl; 
 
 
 
\} 
return 0; 
\}