pat1080. Graduate Admission (30)

时间：2015-09-11 22:06:19 阅读：242 评论：0 收藏：0 [点我收藏+]

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1080. Graduate Admission (30)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

It is said that in 2013, there were 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 G_E, and the interview grade G_I. The final grade of an applicant is (G_E + G_I) / 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 G_E. 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.

Input Specification:

Each input file contains 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 G_E and G_I, 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.

Output Specification:

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.

Sample Input:

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

Sample Output:

提交代码

方法一：

代码改得很久，有些生疏。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<stack>
 5 #include<set>
 6 #include<map>
 7 #include<queue>
 8 #include<algorithm>
 9 using namespace std;
10 struct application{
11     int GE,GI,sum,num;
12     int want[5];//开始写成3，错了！
13 };
14 vector<int> sch[105];
15 application app[40005];
16 int sum[40005],GE[40005];
17 int school[105];
18 bool cmp(application a,application b){
19     if(a.sum==b.sum){
20         return a.GE>b.GE;
21     }
22     return a.sum>b.sum;
23 }
24 int main(){
25     //freopen("D:\\INPUT.txt","r",stdin);
26     int n,m,k,i,j;
27     scanf("%d %d %d",&n,&m,&k);
28     for(i=0;i<m;i++){
29         scanf("%d",&school[i]);
30     }
31     for(i=0;i<n;i++){
32         scanf("%d %d",&app[i].GE,&app[i].GI);
33         GE[i]=app[i].GE;
34         sum[i]=app[i].sum=app[i].GE+app[i].GI;
35         app[i].num=i;
36         for(j=0;j<k;j++){//开始写成3，错了！
37             scanf("%d",&app[i].want[j]);
38         }
39     }
40     sort(app,app+n,cmp);
41 
42 
43     /*for(i=0;i<n;i++){
44         cout<<app[i].GE<<" "<<app[i].GI<<" "<<app[i].sum<<endl;
45     }*/
46 
47 
48     int schnum;
49     for(i=0;i<n;i++){
50         for(j=0;j<k;j++){
51             schnum=app[i].want[j];
52             if(sch[schnum].size()<school[schnum]||(sum[sch[schnum].back()]==app[i].sum&&GE[sch[schnum].back()]==app[i].GE)){//这里要注意排序后，编号都乱了，编号不再和当前的元素对应
53                 sch[schnum].push_back(app[i].num);
54                 break;
55             }
56         }
57     }
58     vector<int>::iterator it;
59     for(i=0;i<m;i++){
60         if(sch[i].size()){
61             sort(sch[i].begin(),sch[i].end());
62             it=sch[i].begin();
63             printf("%d",*it);
64             it++;
65             for(;it!=sch[i].end();it++){
66                 printf(" %d",*it);
67             }
68         }
69         printf("\n");
70     }
71     return 0;
72 }

方法二：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<stack>
 5 #include<set>
 6 #include<map>
 7 #include<queue>
 8 #include<algorithm>
 9 using namespace std;
10 struct application{
11     int GE,GI,sum,num;
12     int want[5];
13 };
14 vector<int> sch[105];
15 application app[40005];
16 int school[105];
17 bool cmp(application a,application b){
18     if(a.sum==b.sum){
19         return a.GE>b.GE;
20     }
21     return a.sum>b.sum;
22 }
23 bool vis[105];//判断是否之前已经满了
24 int main(){
25     //freopen("D:\\INPUT.txt","r",stdin);
26     int n,m,k,i,j;
27     scanf("%d %d %d",&n,&m,&k);
28     for(i=0;i<m;i++){
29         scanf("%d",&school[i]);
30     }
31     for(i=0;i<n;i++){
32         scanf("%d %d",&app[i].GE,&app[i].GI);
33         app[i].sum=app[i].GE+app[i].GI;
34         app[i].num=i;
35         for(j=0;j<k;j++){
36             scanf("%d",&app[i].want[j]);
37         }
38     }
39     sort(app,app+n,cmp);
40 
41 
42     /*for(i=0;i<n;i++){
43         cout<<app[i].GE<<" "<<app[i].GI<<" "<<app[i].sum<<endl;
44     }*/
45     int count;
46     for(i=0;i<n;){
47         queue<int> q;
48         count=0;
49         for(j=i;j<n;j++){
50             if(app[j].sum==app[i].sum&&app[j].GE==app[i].GE){
51                 q.push(j);
52                 count++;
53             }
54             else{
55                 break;
56             }
57         }
58         i=j;
59         int cur;
60         int schnum;
61         int temp;
62         for(j=0;j<k;j++){
63             memset(vis,false,sizeof(vis));
64             temp=count;
65             while(temp--){
66                 cur=q.front();
67                 q.pop();
68                 schnum=app[cur].want[j];
69                 if(school[schnum]>0||vis[schnum]){//是因为同等级的人“虚”满,还有名额
70                     vis[schnum]=true;
71                     school[schnum]--;
72                     sch[schnum].push_back(app[cur].num);
73                     count--;
74                 }
75                 else{
76                     q.push(cur);
77                 }
78             }
79         }
80     }
81     for(i=0;i<m;i++){
82         if(sch[i].size()){
83             sort(sch[i].begin(),sch[i].end());
84             printf("%d",sch[i][0]);
85         }
86         for(j=1;j<sch[i].size();j++){
87             printf(" %d",sch[i][j]);
88         }
89         printf("\n");
90     }
91     return 0;
92 }

pat1080. Graduate Admission (30)

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原文地址：http://www.cnblogs.com/Deribs4/p/4802159.html

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