Ice_cream‘s world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 600    Accepted Submission(s): 344

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

大意：

题意很简单，刚开始我理解成两个塔之间有一个墙把他们隔开呢T_T

分析：并查集判环  只要在一个集合之中一定会有一块空地

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5
 6 const int maxn = 1005;
 7
 8 int fa[maxn];
 9
10 int find(int x) {
11     if(fa[x] == x) return x;
12     return fa[x] = find(fa[x]);
13 }
14
15 int main() {
16     int n, m;
17     int a, b;
18     while(EOF != scanf("%d %d",&n, &m) ) {
19         for(int i = 0; i < n; i++) fa[i] = i;
20         int cnt = 0;
21         while(m--) {
22             scanf("%d %d",&a, &b);
23             if(find(a) == find(b) ) {
24                 cnt++;
25             } else {
26                 fa[find(a)] = find(b);
27             }
28         }
29         printf("%d\n", cnt);
30     }
31     return 0;
32 }