Ice_cream's world I
Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 1 Accepted Submission(s) : 1
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
简单并差集，求环路数量；
代码：
1 #include<stdio.h> 2 int wall[1010]; 3 int find(int x){ 4 int r=x; 5 while(r!=wall[r])r=wall[r]; 6 int i=x,j; 7 while(i!=r)j=wall[i],wall[i]=r,i=j; 8 return r; 9 } 10 int main(){ 11 int N,M,temp1,temp2,f1,f2,flot; 12 while(~scanf("%d%d",&N,&M)){ 13 for(int i=0;i<N;++i)wall[i]=i; 14 flot=0; 15 while(M--){ 16 scanf("%d%d",&temp1,&temp2); 17 f1=find(temp1);f2=find(temp2); 18 if(f1!=f2)wall[f1]=f2; 19 else flot++; 20 } 21 printf("%d ",flot); 22 } 23 return 0; 24 }