HDOJ 2120 并查集

并查集的应用，用来查找被分割的区域个数。

即当两个节点值相同时说明已经为了一个圈，否则不可能，此时区域个数加1.

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 const int maxn=1010;
 6 int n,m;
 7 int root[maxn];
 8 
 9 int find(int a){
10     while(root[a]!=a){
11         a=root[a];
12     }
13     return a;
14 }
15 
16 int main(){
17     while(EOF != scanf("%d%d",&n,&m)){
18         for(int i=0;i<n;i++){
19             root[i]=i;
20         }
21         int ans = 0;
22         int a,b;
23         while(m--){
24             scanf("%d%d",&a,&b);
25             int ra=find(a);
26             int rb=find(b);
27             if(ra == rb)
28                 ans++;
29             else
30                 root[ra]=rb;//epual to Join
31             //printf("%d %d %d
",ra,rb,ans);
32         }
33         printf("%d
",ans);
34     }
35     return 0;
36 }

Ice_cream's world I

Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 3 Accepted Submission(s) : 2

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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