数据结构(陈越 、何钦铭)--第五周编程作业

05-树7 堆中的路径 (25分)

将一系列给定数字插入一个初始为空的小顶堆H[]。随后对任意给定的下标i，打印从H[i]到根结点的路径。

输入格式:
每组测试第1行包含2个正整数N和M(≤1000)，分别是插入元素的个数、以及需要打印的路径条数。下一行给出区间[-10000, 10000]内的N个要被插入一个初始为空的小顶堆的整数。最后一行给出M个下标。

输出格式:
对输入中给出的每个下标i，在一行中输出从H[i]到根结点的路径上的数据。数字间以1个空格分隔，行末不得有多余空格。

输入样例:
5 3
46 23 26 24 10
5 4 3
输出样例:
24 23 10
46 23 10
26 10

#include 
#define MAXSIZE 1001

int H[MAXSIZE];

void CreatS(int N, int H[]);
void Output(int M, int H[]);

int main()
{
    int N, M;
    scanf("%d %d", &N, &M);

    CreatS(N, H);
    Output(M, H);

    return 0;
}

void CreatS(int N, int H[])
{
    H[0] = -10001;
    int i, j, item;

    for(i = 1; i <= N; i++){
        scanf("%d", &item);
        for(j = i; H[j/2] > item; j/=2){
            H[j] = H[j/2];
        }
        H[j] = item;
    }
}

void Output(int M, int H[])
{
    int i, item, j;
    for(i = 0; i < M; i++){
        int flag = 0;
        scanf("%d", &item);
        for(j = item; j > 0; j/=2){
            if(flag){
                printf(" ");
            }
            printf("%d", H[j]);
            flag = 1;
        }
        printf("\n");
    }
}

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≤10^4), 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.

#include 
#define MAXSIZE 10000

typedef int ElementType;
typedef int SetName;

void ConnectCom(ElementType Group[]);
void CheckConnection(ElementType Group[]);
SetName FindRoot(SetName a, ElementType Group[]);

int main()
{
    ElementType Group[MAXSIZE];
    int N, i;
    char in;
    scanf("%d", &N);
    for(i = 0; i < N; i++){
        Group[i] = -1;
    }

    do{
        scanf("%c", &in);
        switch(in)
        {
            case 'I':
                ConnectCom(Group);    break;
            case 'C':
                CheckConnection(Group);    break;
        }
    }while(in != 'S');

    int min = 0;
    for(i = 0; i < N; i++){
        if(Group[i] < 0)    min++;
    }
    if(min == 1){
        printf("The network is connected.");
    }else{
        printf("There are %d components.", min);
    }

    return 0;
}

SetName FindRoot(SetName a, ElementType Group[])
{
    if(Group[a] < 0)
        return a;
    else
        return Group[a] = FindRoot(Group[a], Group);
}

void ConnectCom(ElementType Group[])
{
    SetName a, b;
    scanf("%d %d", &a, &b);
    a = FindRoot(a, Group);
    b = FindRoot(b, Group);
    if(Group[a] <= Group[b]){
        Group[a] += Group[b];
        Group[b] = a;
    }else{
        Group[b] += Group[a];
        Group[a] = b;
    }
}

void CheckConnection(ElementType Group[])
{
    SetName a, b;
    scanf("%d %d", &a, &b);
    a = FindRoot(a, Group);
    b = FindRoot(b, Group);
    if(a == b)    printf("yes\n");
    else    printf("no\n");
}