05-树8 File Transfer (25分)
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题目描述
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≤10000), 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.
解题思路
这道题是典型的并查集问题,我们可以用树来表示每一个集合,用树的根结点值来代表集合名,又根据这道题的输入数据,我们可以用一个数组实现这些树,数组的下标对应电脑标号减1,元素值为其父结点的下标,根结点的值为-结点数。由题可知,需要处理I
,C
,S
三种情况。对于I
来说,我们首先需要得到两个数据对应的集合,若不同则合并,若相同则不管。于是这就可以抽象成两个问题,即得到数据对应的集合与合并两个集合。对于C
来说,我们只需得到两个数据对应的集合,若相同打印yes,否则打印no。对于S
来说,我们只需遍历数组,统计值小于0的元素个数,然后根据情况打印不同的字符串。
代码
#include <stdio.h>
#define MAXSIZE 10000
typedef int SetName; //用根结点的值表示集合
typedef int SetType[MAXSIZE]; //用一个数组表示集合,下标是电脑标号减一,值是父结点的下标
void init(SetType set, int N);
SetName getSetName(SetType set, int v);
void merge(SetType set, SetName root1, SetName root2);
void checkConn(SetType set);
void connection(SetType set);
void checkSet(SetType set, int N);
int main() {
int N;
char op;
SetType set;
scanf("%d
", &N);
init(set, N);
do {
scanf("%c", &op);
switch (op) {
case ‘C‘:
checkConn(set); break;
case ‘I‘:
connection(set); break;
case ‘S‘:
checkSet(set, N); break;
}
} while (op != ‘S‘);
return 0;
}
//将数组前N个数初始化为-1
void init(SetType set, int N) {
for (int i = 0; i < N; i++) {
set[i] = -1;
}
}
//得到下标v对应的集合名
SetName getSetName(SetType set, int v) {
for ( ; set[v] >= 0; v = set[v]);
return v;
}
//求两个集合的并集,为了尽可能地降低集合高度,将结点数小的集合的根连到结点数大的集合的根
void merge(SetType set, SetName root1, SetName root2) {
if (set[root1] < set[root2]) { //集合1结点更多
set[root1] += set[root2];
set[root2] = root1;
} else {
set[root2] += set[root1];
set[root1] = root2;
}
}
//判断两个值是否属于同一集合
void checkConn(SetType set) {
int a, b;
scanf("%d %d
", &a, &b);
SetName root1, root2;
root1 = getSetName(set, a - 1);
root2 = getSetName(set, b - 1);
if (root1 == root2) printf("yes
");
else printf("no
");
}
//将两个值对应的集合并起来
void connection(SetType set) {
int a, b;
scanf("%d %d
", &a, &b);
SetName root1, root2;
root1 = getSetName(set, a - 1);
root2 = getSetName(set, b - 1);
if (root1 != root2) merge(set, root1, root2);
}
//统计根结点个数,根据情况输出
void checkSet(SetType set, int N) {
int count = 0;
for (int i = 0; i < N; i++) {
if (set[i] < 0) count++;
}
if (count == 1) printf("The network is connected.
");
else printf("There are %d components.
", count);
}
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