九度OJ 1005：Graduate Admission （排序）
- 题目描述：
-
It is said that in 2011, there are about 100 graduate schools ready to proceed over 40,000 applications in Zhejiang Province. It would help a lot if you could write a program to automate the admission procedure.
Each applicant will have to provide two grades: the national entrance exam grade GE, and the interview grade GI. The final grade of an applicant is (GE + GI) / 2. The admission rules are:• The applicants are ranked according to their final grades, and will be admitted one by one from the top of the rank list.
• If there is a tied final grade, the applicants will be ranked according to their national entrance exam grade GE. If still tied, their ranks must be the same.
• Each applicant may have K choices and the admission will be done according to his/her choices: if according to the rank list, it is one's turn to be admitted; and if the quota of one's most preferred shcool is not exceeded, then one will be admitted to this school, or one's other choices will be considered one by one in order. If one gets rejected by all of preferred schools, then this unfortunate applicant will be rejected.
• If there is a tied rank, and if the corresponding applicants are applying to the same school, then that school must admit all the applicants with the same rank, even if its quota will be exceeded.
- 输入：
-
Each input file may contain more than one test case.
Each case starts with a line containing three positive integers: N (≤40,000), the total number of applicants; M (≤100), the total number of graduate schools; and K (≤5), the number of choices an applicant may have.
In the next line, separated by a space, there are M positive integers. The i-th integer is the quota of the i-th graduate school respectively.
Then N lines follow, each contains 2+K integers separated by a space. The first 2 integers are the applicant's GE and GI, respectively. The next K integers represent the preferred schools. For the sake of simplicity, we assume that the schools are numbered from 0 to M-1, and the applicants are numbered from 0 to N-1.
- 输出：
-
For each test case you should output the admission results for all the graduate schools. The results of each school must occupy a line, which contains the applicants' numbers that school admits. The numbers must be in increasing order and be separated by a space. There must be no extra space at the end of each line. If no applicant is admitted by a school, you must output an empty line correspondingly.
- 样例输入：
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11 6 3 2 1 2 2 2 3 100 100 0 1 2 60 60 2 3 5 100 90 0 3 4 90 100 1 2 0 90 90 5 1 3 80 90 1 0 2 80 80 0 1 2 80 80 0 1 2 80 70 1 3 2 70 80 1 2 3 100 100 0 2 4
- 样例输出：
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0 10 3 5 6 7 2 8 1 4
思路：
本题考查的是排序，其中有很多坑的地方，我在做的过程中出了很多错误，看论坛参考别人意见才一一发现解决。
其中最可能坑的地方是：先给考生排名后，根据考生排名从名次高到名次低逐个录取，考生第一志愿没录取直接看第二志愿，跟我们高考不一样的。
另外还需要注意的：当学校没有已经没有名额时，若你与所报的学校已经录取的最低分相等且为同等志愿，你就会破格被录取。。。（也就是为什么样例中的学生7会被学校2录取）
数据结构用结构体比较好，用分开的数组就太乱了。C语言用qsort排序，C++用sort。
代码：
#include <stdio.h>
#include <stdlib.h>
#define N 40000
#define M 100
#define K 5
typedef struct student {
int id;
int ge;
int gi;
int sch[K];
} Stu;
typedef struct school {
int quota;
int num;
int ge;
int gi;
int stuid[N];
} Sch;
int n, m, k;
Stu stu[N];
Sch sch[M];
int cmpGrade(const void *a, const void *b)
{
Stu *x = (Stu *)a;
Stu *y = (Stu *)b;
if (x->ge + x->gi != y->ge + y->gi)
return (y->ge + y->gi) - (x->ge + x->gi);
else
return y->ge - x->ge;
}
int cmpId(const void *a, const void *b)
{
return *(int *)a - *(int *)b;
}
void printSch(int i)
{
int j;
int num = sch[i].num;
for (j=0; j<num-1; j++)
printf("%d ", sch[i].stuid[j]);
if (num > 0)
printf("%d\n", sch[i].stuid[j]);
else
printf("\n");
}
int main(void)
{
int i, j;
while (scanf("%d%d%d", &n, &m, &k) != EOF)
{
for (i=0; i<m; i++)
{
scanf("%d", &sch[i].quota);
sch[i].num = 0;
}
for (i=0; i<n; i++)
{
scanf("%d%d", &stu[i].ge, &stu[i].gi);
stu[i].id = i;
for (j=0; j<k; j++)
{
scanf("%d", &(stu[i].sch[j]));
}
}
qsort(stu, n, sizeof(stu[0]), cmpGrade);
for (i=0; i<n; i++)
{
for (j=0; j<k; j++)
{
int schid = stu[i].sch[j];
int num = sch[schid].num;
if ( (num < sch[schid].quota) || ( (num >= sch[schid].quota)
&& (stu[i].ge == sch[schid].ge)
&& (stu[i].gi == sch[schid].gi) ) )
{
sch[schid].stuid[num] = stu[i].id;
sch[schid].ge = stu[i].ge;
sch[schid].gi = stu[i].gi;
sch[schid].num ++;
//printf("schid=%d, num=%d\n", schid, num);
break;
}
}
}
for (i=0; i<m; i++)
{
qsort(sch[i].stuid, sch[i].num, sizeof(int), cmpId);
printSch(i);
}
}
return 0;
}
/**************************************************************
Problem: 1005
User: liangrx06
Language: C
Result: Accepted
Time:0 ms
Memory:17792 kb
****************************************************************/
