Ice_cream's world I
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 758 Accepted Submission(s): 442
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
Author
Wiskey
Source
Recommend
//计算成环个数;
1 #include <stdio.h> 2 int father[1010]; 3 int find(int a) 4 { 5 while(a != father[a]) 6 a = father[a]; 7 return a; 8 } 9 void mercy(int a, int b) 10 { 11 int q = find(a); 12 int p = find(b); 13 if(q != p) 14 father[q] = p; 15 } 16 int main() 17 { 18 int i, n, m; 19 while(~scanf("%d %d", &n, &m)) 20 { 21 int total = 0; 22 for(i=0; i<n; i++) 23 father[i] = i; 24 for(i=0; i<m; i++) 25 { 26 int a, b; 27 scanf("%d %d", &a, &b); 28 if(find(a) == find(b)) 29 { 30 //printf(" %d %d ", a, b); 31 total++; 32 } 33 else 34 mercy(a, b); 35 } 36 printf("%d ", total); 37 } 38 }