c语言实训题目设计报告模拟高考平行志愿投档,（求）C语言：模拟高考平行志愿投档...

偏执的太偏执、 2022-10-14 05:53 162阅读 0赞

\#include

typedef struct struct\_node\{

int number;

int score;

int application\[3\];

struct struct\_node \* next;

\} node;

typedef struct \{

node \* first;

node \* last;

\} queue;

typedef struct \{

node \* head;

\} list;

void init\_queue(queue \* q) \{

q->first = q->last = NULL;

void enqueue(queue \* q, node \* n) \{

if (q->first == NULL) \{

q->first = q->last = n;

\} else \{

q->last->next = n;

q->last = q->last->next;

node \* dequeue(queue \* q) \{

node \* n = q->first;

if (q->first == q->last) \{

q->first = q->last = NULL;

\} else \{

q->first = q->first->next;

return n;

int init\_list(list \* l) \{

l->head = (node \*) malloc(sizeof(node));

if (l->head) \{

l->head->next = NULL;

return 0;

\} else \{

return 1;

void insert\_list\_element(list \* l, node \* n) \{

node \* q = l->head, \* p = q->next;

while (p && p->score > n->score) \{

q = p;

p = p->next;

q->next = n;

n->next = p;

void remove\_list\_element(list \* l, node \* n) \{

node \* p = l->head;

if (n == NULL) \{

return;

while (p && p->next != n) \{

p = p->next;

if (p) \{

p->next = p->next->next;

int main() \{

int m, \* need, n, i;

list applicants;

node \* applicant, \* p;

queue \* position;

printf("\\nHow many positions available are there?");

fflush(stdout);

scanf("%d", &m);

need = (int \*) malloc(sizeof(int) \* m);

if (need == NULL) \{

printf("\\nFailed to allocate memory.\\nApplication quited.\\n");

fflush(stdout);

return 1;

printf("\\nHow many people are needed for every position?\\n");

printf("Flow the order position 1, position 2 ...\\n");

fflush(stdout);

for (i = 0; i < m; i++) \{

scanf("%d", &need\[i\]);

printf("\\nHow many applicants are there?");

fflush(stdout);

scanf("%d", &n);

if (init\_list(&applicants)) \{

printf("\\nFailed to allocate memory.\\nApplication quited.\\n");

fflush(stdout);

return 1;

printf("\\nTell me about the applicants in this format:\\n");

printf("score application\_1 application\_2\\n");

printf("Following the order applicant 1, applicant 2 ...\\n");

fflush(stdout);

for (i = 0; i < n; i++) \{

applicant = (node \*) malloc(sizeof(node));

applicant->number = i;

scanf("%d %d %d", &applicant->score,

&applicant->application\[1\], &applicant->application\[2\]);

applicant->application\[0\] = applicant->application\[1\];

insert\_list\_element(&applicants, applicant);

position = (queue \*) malloc(sizeof(queue) \* m);

if (position == NULL) \{

printf("\\nFailed to allocate memory.\\nApplication quited.\\n");

return 1;

for (i = 0; i < m; i++) \{

init\_queue(&position\[i\]);

//Employ the applicants according to their scores.//Let p point at the head of list applicants. Couse the list applicants is sorted descendingly,//at beginning, p->next points at the applicant with the best score. for (p = applicants.head, applicant = p->next; applicant; applicant = p->next) \{

//If the position for which the applicant applied is available. if (need\[applicant->application\[0\]\]) \{

//Employe the applicant to the position he/she applied for. enqueue(&position\[applicant->application\[0\]\], applicant);

//Couse the applicant is employed already, he/she should get out of the list of applicants. remove\_list\_element(&applicants, applicant);

//Couse we employd 1 person for the position, the need of the position should be decreased by 1. need\[applicant->application\[0\]\]--;

//If the position for which the applicant applied is NOT available. \} else \{

//If the application is the applicant's second one, he/she can not be employed if (applicant->application\[0\] == applicant->application\[2\]) \{

//Couse his/her score was subtracted by 5 when we try to employ him/her according to his/her second//application, now we need to restore his/her score to original state. applicant->score += 5;

//Let p point at the applicant. That means, from now on, wo employ only the people behind him/her,//all the people and him/her will never be consider any more. p = applicant;

//If the application is the applicant's first one, he/she still has chance to be employed \} else \{

//We subtract the applicant's score by 5. applicant->score -= 5;

//Try to employ him/her according to his/her second appliction. applicant->application\[0\] = applicant->application\[2\];

//By let him/her out of the list and comback again, he/she is arranged to the correct position//in the according to his/her new score. remove\_list\_element(&applicants, applicant);

insert\_list\_element(&applicants, applicant);

//Show the employed applicants. for (i = 0; i < m; i++) \{

printf("\\nThe applicants employed for position %d are:\\n", i);

while (position\[i\].first) \{

applicant = dequeue(&position\[i\]);

printf("No.%d Score: %2d 1st application: %d 2nd application: %d\\n",

applicant->number, applicant->score, applicant->application\[1\], applicant->application\[2\]);

fflush(stdout);

applicant = applicants.head->next;

if (applicant == NULL) \{

printf("\\nAll applicants are employed.\\n");

\} else \{