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.
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 ", cnt); 30 } 31 return 0; 32 }