这个按照老师的自己码一遍就好了,按秩合并+路径压缩,然后最后的K是指在整个网络里有多少个集合。不说理论部分,不相关集是我本科数据结构学的代码最简单的部分了,哈哈。
05-树8 File Transfer(25 分)
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:
5
C 3 2
I 3 2
C 1 5
I 4 5
I 2 4
C 3 5
S
Sample Output 1:
no
no
yes
There are 2 components.
Sample Input 2:
5
C 3 2
I 3 2
C 1 5
I 4 5
I 2 4
C 3 5
I 1 3
C 1 5
S
Sample Output 2:
no
no
yes
yes
The network is connected.
/*-------------------------------------------------------------- 如果 ElementType Data是个整数的话,那么就直接用数组好了 ---------------------------------------------------------------*/ #include <stdio.h> #define MaxSize 100000 typedef int ElementType; typedef int SetName; typedef ElementType SetType[MaxSize]; SetName Find(SetType S, ElementType X); void Union (SetType S, SetName Root1, SetName Root2); void Input_connections(SetType S); /* 路径压缩,这个可以通过第四个测试点*/ SetName Find(SetType S, ElementType X){ if(S[X] < 0) return X; else return S[X] = Find(S, S[X]); } void Check_network(SetType S, int n){ int i, counter = 0; for(i = 0; i < n; i++){ if(S[i] < 0) counter++; } if(counter == 1) printf("The network is connected.\n"); else printf("There are %d components.\n",counter); } void Input_connections(SetType S){ ElementType u,v; SetName Root1, Root2; scanf("%d %d\n",&u, &v); Root1 = Find(S, u-1); Root2 = Find(S, v-1); if(Root1 != Root2) Union(S, Root1, Root2); } void Check_connections(SetType S){ ElementType u,v; SetName Root1,Root2; scanf("%d %d\n",&u,&v); Root1 = Find(S, u-1); Root2 = Find(S, u-1); if(Root1 == Root2) printf("yes\n"); else printf("no\n"); } /* 按秩归并, 如果随意归并的话很可能会超时, 就是把矮树贴到高的树上,用根节点表示树高,这样可以past一个测试点 */ void Union (SetType S, SetName Root1, SetName Root2){ if(S[Root2] < S[Root1]) S[Root1] = Root2; //小的连在大的,二者的高度都不会变; else{ if(S[Root2] == Root1) S[Root1] --; //相等的话,选择其中一个改变,这里选择root1; S[Root2] = Root1; } } void Initialization(SetType S, int n){ for (int i = 0; i < n; i++) { S[i] = -1; } } int main(int argc, const char * argv[]) { SetType S; int N; scanf("%d",&N); Initialization(S,N); char in; do{ scanf("%c",&in); switch (in) { case 'I': Input_connections(S); break; case 'C': Check_connections(S); break; case 'S': Check_network(S,N); break; } }while(in != 'S'); return 0; }