Ice_cream's world I(并查集成环)

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   Source HDU 2007-10 Programming Contest_WarmUp  分析:并查集成环问题。。关于并查集理解了为什么能判断成环,首先并查集我们知道能够判断两个点是否属于同一个集合,并且将不属于同一个集合中的点归纳到同一个集合中去。。。 Ice_cream's world I(并查集成环)

这个图有环是已知的吧,但是要用代码来判断,就很复杂,首先我们把每个点看成独立的集合{0} ,{1}, {2}, 然后规定如果两个点之间有边相连,如果这两个点不属于同一个集合,那就将他们所属的结合合并,看边0-1,直接将这两个点代表的集合合并{0, 1}, 其中让1来当父节点, 看边1-2, 它们分别属于不同的集合,合并集合之后是{1, 2},让2来当父节点,依照这种逻辑关系,0的祖先节点就是2, 然后在看边0-2,他们属于一个集合,因为他们有着共同的祖先2,
这就说明0-2之间在没有0-2这条边之前已经连通了,如果在加上这条边的话那从0到2就有两条路径可达,就说明存在一个环了。。。这就是并查集所谓的成环的实质。。。

 1 // /*并查集*/
 2 // int prev[1000];
 3 
 4 // int find(int x){//查找我的掌门
 5 //     int r=x;    //委托r去找掌门
 6 //     while(prev[r]!=r){//如果r的上级不是自己(也就是说他找到的大侠不是掌门)
 7 //         r=prev[r];//r就接着找他的掌门,直到到掌门为止
 8 //     }
 9 //     return r;//掌门驾到~~~~
10 // }
11 
12 // void join(int x,int y){//联通
13 //     int fx=find(x);
14 //     int fy=find(y);
15 
16 //     if(fx!=fy){
17 //         prev[fx]=fy;
18 //     }
19 
20 // }
21 #include<iostream>
22 #include<cstdio>
23 #include<cstring>
24 #include<algorithm>
25 using namespace std;
26 
27 const int maxn = 1e3+10;
28 int pre[maxn];
29 int cnt=0;
30 
31 int Find(int x){
32     int u=x;
33     while(u!=pre[u]){
34         u=pre[u];
35     }
36     int i=x,j;
37     while(pre[i]!=u){/*压缩路径*/
38         j=pre[i];
39         pre[i]=u;
40         i=j;
41     }
42     return u;
43 }
44 
45 void mix(int u,int v){
46     int fu=Find(u);
47     int fv=Find(v);
48     if(fu!=fv){
49         pre[fu]=fv;
50     }
51     else{
52         cnt++;
53     }
54 }
55 
56 int main(int argc, char const *argv[])
57 {
58     int n,m;
59      while(~scanf("%d %d",&n,&m)){
60          cnt=0;
61         for( int i=0; i<n; i++ ){/*初始化*/
62             pre[i]=i;
63         }
64         for( int i=0; i<m; i++ ){
65             int a,b;
66             cin>>a>>b;
67             mix(a,b);
68         }
69         cout<<cnt<<endl;
70     }
71     return 0;
72 }

 

上一篇:多态


下一篇:IceGrid应用 配置手册