hdu1962Corporative Network带权回路

简介:
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 1 /*
 2     有N个企业,每个企业想要实现通信,要用线路来连接,线路的长度为abs(a-b)%1000;
 3     如果企业a 链接到了企业b 那么b就是the center of the serving!
 4     然后有两种操作:
 5     E a : 输出企业a到serving center 的线路的距离
 6     I a, b  将企业a连接到企业 b,那么b就成为了serving center(之前连接a的企业,他们的serving center也变成了b) 
 7     
 8    思路:并查集! (压缩路径时回溯求解) ! 
 9 */ 
10 #include<iostream>
11 #include<cstring>
12 #include<cmath>
13 #include<cstdio>
14 #define M 20005
15 using namespace std;
16 int n;
17 int f[M];
18 int ans[M];//节点 i到 serving center的距离! 
19 
20 int getFather(int x){
21     if(x==f[x]) return x;
22     int ff=getFather(f[x]);
23     ans[x]+=ans[f[x]];//节点x到serving center 的距离要加上其父节点到serving center的距离! 
24     return f[x]=ff;
25 }
26 
27 void Union(int a, int b){ 
28     if(a==b) return;
29     f[a]=b;
30     ans[a]=abs(a-b) % 1000;
31 }
32 
33 int main(){
34    int t;
35    char ch[3];
36    cin>>t;
37    while(t--){
38       cin>>n;
39       int a, b;
40       memset(ans, 0, sizeof(ans));
41       for(int i=1; i<=n; ++i)
42          f[i]=i;
43       while(cin>>ch && ch[0]!='O'){
44           if(ch[0]=='E'){
45              cin>>a;
46              getFather(a);
47              cout<<ans[a]<<endl;
48           }
49           else{
50              cin>>a>>b;
51              Union(a, b);
52           }
53       }
54    }
55    return 0;
56 }









本文转自 小眼儿 博客园博客,原文链接:http://www.cnblogs.com/hujunzheng/p/3902548.html,如需转载请自行联系原作者
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