04-树5. File Transfer (25)

We have a network of computers and a list of bi-directional connections. Each of these connections allows a file transfer from one computer to another. Is it possible to send a file from any computer on the network to any other?

Input Specification:

Each input file contains one test case. For each test case, the first line contains N (2<=N<=10⁴), the total number of computers in a network. Each computer in the network is then represented by a positive integer between 1 and N. Then in the following lines, the input is given in the format:

I c1 c2  

where I stands for inputting a connection between c1 and c2; or

C c1 c2    

where C stands for checking if it is possible to transfer files between c1 and c2; or

S

where S stands for stopping this case.

Output Specification:

For each C case, print in one line the word "yes" or "no" if it is possible or impossible to transfer files between c1 and c2, respectively. At the end of each case, print in one line "The network is connected." if there is a path between any pair of computers; or "There are k components." where k is the number of connected components in this network.

Sample Input 1:

5C 3 2I 3 2C 1 5I 4 5I 2 4C 3 5S

Sample Output 1:

nonoyesThere are 2 components.

Sample Input 2:

5C 3 2I 3 2C 1 5I 4 5I 2 4C 3 5I 1 3C 1 5S

Sample Output 2:

nonoyesyesThe network is connected.

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这个是并集的问题，在网易云课堂上面，老师已经讲过了。

主要还是找父亲（root）的问题，这里用到的是数组，直接用数组来表示其root。

刚刚开始的时候，初始化root为其本身，

然后遇到“I”的时候合并。

这里还有一个问题就是没有考虑集合的高度，会导致算法的耗时，有一定的提升的空间，就查了下其他同学的，一并贴在下面。

代码如下。

#include <iostream>using namespace std;/* run this program using the console pauser or add your own getch, system("pause") or input loop */int fa[10002];int N=0;int findfa(int a){if(a != fa[a])fa[a]=findfa(fa[a]);return fa[a];}void add(int a, int b){if(findfa(a) != findfa(b))//not connected{fa[a] = b;//it is not effective! The height should be considered!N--;}}int main(int argc, char** argv) {int c1=0,c2=0;char ch;cin>>N;for(int i=1; i<=N;i++){fa[i]=i;}cin>>ch;while(ch != 'S'){cin>>c1>>c2;if(ch == 'C'){if(findfa(c1) == findfa(c2)) printf("yes\n");else printf("no\n");}else if(ch == 'I'){add(c1,c2);}cin>>ch;}//The connected computersif(N==1)printf("The network is connected.\n");elseprintf("There are %d components.\n",N);return 0;}

有位同学（http://blog.csdn.net/u013167299/article/details/42243501），考虑到了树的高度问题，这样会在耗时上比自己的有优势。

代码如下。

学习一下。

#include <iostream>  #include <algorithm>  #include <string>  #include <cstdio>  #include <cstring>  #include <cstdlib>  #include <cmath>  #include <vector>  #include<queue>  #include<stack>  #include<map>  #include<set>  using namespace std;  #define lson rt<<1,l,MID  #define rson rt<<1|1,MID+1,r  //#define lson root<<1  //#define rson root<<1|1  #define MID ((l+r)>>1)  typedef long long ll;  typedef pair<int,int> P;  const int maxn=50005;  const int base=1000;  const int inf=999999;  const double eps=1e-5;  int a[maxn];//表示集合的树  int r[maxn];//表示树的高度  void build(int n)//初始化  {      for(int i=0;i<=n;i++)      {          a[i]=i;//父亲就是自己          r[i]=0;//高度为0       }  }    int find(int x)//查找节点的父亲  {      if(a[x]==x)          return x;      return a[x]=find(a[x]);  }    void add(int x,int y)//节点的合并  {      x=find(a[x]);      y=find(a[y]);      if(x==y)          return;      else if(r[x]<r[y])          a[x]=y;      else      {          a[y]=x;          if(r[x]==r[y])r[x]++;      }  }    bool same(int x,int y)//判断是不是一个集合  {      return find(x)==find(y);  }    int main()  {      int n,m,i,j,k,t;      cin>>n;      build(n);      while(1)      {          char op[2];          cin>>op;          if(op[0]=='I')          {              cin>>i>>j;              add(i,j);          }          else if(op[0]=='C')          {              cin>>i>>j;              if(same(i,j))                  puts("yes");              else                  puts("no");          }          else if(op[0]=='S')          {              int cnt=0;              for(i=1;i<=n;i++)                  if(i==a[i])cnt++;              if(cnt<=1)                  puts("The network is connected.");              else                  printf("There are %d components.\n",cnt);              break;          }      }      return 0;  }