hdu2120 Ice_cream's world I (并查集查找环的个数)

hdu2120 Ice_cream's world I (并查集查找环的个数)

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 887    Accepted Submission(s): 517

Problem Description

ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

Input

In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.

Output

Output the maximum number of ACMers who will be awarded. One answer one line.

Sample Input

8 10 0 1 1 2 1 3 2 4 3 4 0 5 5 6 6 7 3 6 4 7

Sample Output

3

Author

Wiskey

Source

​​HDU 2007-10 Programming Contest_WarmUp​​

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​​Note​​

这道题的意思就是查找一个图里面有几个环。

怪我一切知识都学的太死了。唉，起初竟然没想到。。多希望再来几个这样的题

让我捡漏。。

#include int fa[1005];int find(int x){ if(fa[x]!=x) fa[x]=find(fa[x]); return fa[x];}bool comb(int a,int b){ a=find(a); b=find(b); if(a==b)//如果根相等就是出现了环 return true; else { fa[a]=b; return false; }}void init(int n){ for(int i=0;i

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