Ice_cream's world I

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\n",flot);
22     }
23     return 0;
24 }

 

posted @ 2015-07-09 09:34  handsomecui  阅读(211)  评论(0编辑  收藏  举报