pat1012. The Best Rank (25)

1012. The Best Rank (25)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algebra), and E - English. At the mean time, we encourage students by emphasizing on their best ranks -- that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.

For example, The grades of C, M, E and A - Average of 4 students are given as the following:

StudentID  C  M  E  A
310101     98 85 88 90
310102     70 95 88 84
310103     82 87 94 88
310104     91 91 91 91

Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.

Input

Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (<=2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.

Output

For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.

The priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.

If a student is not on the grading list, simply output "N/A".

Sample Input 
5 6
310101 98 85 88
310102 70 95 88
310103 82 87 94
310104 91 91 91
310105 85 90 90
310101
310102
310103
310104
310105
999999
Sample Output 
1 C
1 M
1 E
1 A
3 A
N/A

提交代码   

 

注意点：

1.题目先给出n个学生的情况，先对这n个学生按照A、C、M、E的顺序依次进行排序，按照题意得到每个学生最好的排名情况，存储下来。然后，查询m个学生，如果学生不存在，则输出“N/A”，否则，输出最好排名及对应A、C、M、E的哪种情况。

2.题目中如果排名是1 2 2 4 5 6 6 8规则排名（见代码），却不是1 2 2 3 4 5 5 6 

3.map的使用：

http://blog.sina.com.cn/s/blog_61533c9b0100fa7w.html

 

添加元素：

map<int ,string> maplive;  

1）maplive.insert(pair<int,string>(102,"aclive"));
2）maplive.insert(map<int,string>::value_type(321,"hai"));
3）maplive[112]="April";                      //map中最简单最常用的插入添加！

 

查找元素：

find()函数返回一个迭代器指向键值为key的元素，如果没找到就返回指向map尾部的迭代器。        

map<int ,string >::iterator l_it;; 
l_it=maplive.find(112);
if(l_it==maplive.end())
                cout<<"we do not find 112"<<endl;
else cout<<"wo find 112"<<endl;

 

删除元素：

map<int ,string >::iterator l_it;
l_it=maplive.find(112);
if(l_it==maplive.end())
        cout<<"we do not find 112"<<endl;
else  maplive.erase(l_it);    //删除元素

 

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <string>
  4 #include <queue>
  5 #include <stack>
  6 #include <algorithm>
  7 #include <map>
  8 #include <iostream>
  9 using namespace std;
 10 struct cnode{
 11     int grade;
 12     string id;
 13 };
 14 struct mnode{
 15     int grade;
 16     string id;
 17 };
 18 struct enode{
 19     int grade;
 20     string id;
 21 };
 22 struct anode{
 23     int grade;
 24     string id;
 25 };
 26 struct somenode{
 27     int rank;
 28     string r;
 29 };
 30 map<string,somenode> some;
 31 bool cmpa(anode a,anode b){
 32     return a.grade>b.grade;
 33 }
 34 bool cmpc(cnode a,cnode b){
 35     return a.grade>b.grade;
 36 }
 37 bool cmpm(mnode a,mnode b){
 38     return a.grade>b.grade;
 39 }
 40 bool cmpe(enode a,enode b){
 41     return a.grade>b.grade;
 42 }
 43 int main(){
 44     //freopen("D:\\INPUT.txt","r",stdin);
 45     int n,m,cg,mg,eg;
 46     string num;
 47     scanf("%d %d",&n,&m);
 48     anode *ap=new anode[n];
 49     cnode *cp=new cnode[n];
 50     mnode *mp=new mnode[n];
 51     enode *ep=new enode[n];
 52     int i;
 53     for(i=0;i<n;i++){
 54         cin>>num;
 55         scanf("%d %d %d",&cg,&mg,&eg);
 56         ap[i].id=ep[i].id=cp[i].id=mp[i].id=num;
 57         ap[i].grade=int((cg+mg+eg)*1.0/3+0.5);      //四舍五入
 58         ep[i].grade=eg;
 59         mp[i].grade=mg;
 60         cp[i].grade=cg;
 61     }
 62 
 63 
 64     sort(ap,ap+n,cmpa);
 65     int cal=1;
 66     somenode p;
 67     p.r="A";
 68     p.rank=cal;
 69     some[ap[0].id]=p;
 70     for(i=1;i<n;i++){
 71             if(ap[i].grade!=ap[i-1].grade){
 72                 cal=i+1;
 73             }
 74             p.r="A";
 75             p.rank=cal;
 76             some[ap[i].id]=p;
 77     }
 78 
 79 
 80     sort(cp,cp+n,cmpc);
 81     cal=1;
 82     if(some[cp[0].id].rank>cal){
 83         some[cp[0].id].rank=cal;
 84         some[cp[0].id].r="C";
 85     }
 86     for(i=1;i<n;i++){
 87         if(cp[i].grade!=cp[i-1].grade){
 88                 cal=i+1;
 89         }
 90         if(some[cp[i].id].rank>cal){
 91            some[cp[i].id].rank=cal;
 92            some[cp[i].id].r="C";
 93         }
 94     }
 95 
 96 
 97     sort(mp,mp+n,cmpm);
 98     cal=1;
 99     if(some[mp[0].id].rank>cal){
100         some[mp[0].id].rank=cal;
101         some[mp[0].id].r="M";
102     }
103     for(i=1;i<n;i++){
104         if(mp[i].grade!=mp[i-1].grade){
105                 cal=i+1;
106         }
107         if(some[mp[i].id].rank>cal){
108            some[mp[i].id].rank=cal;
109            some[mp[i].id].r="M";
110         }
111     }
112 
113 
114     sort(ep,ep+n,cmpe);
115     cal=1;
116     if(some[ep[0].id].rank>cal){
117         some[ep[0].id].rank=cal;
118         some[ep[0].id].r="E";
119     }
120     for(i=1;i<n;i++){
121         if(ep[i].grade!=ep[i-1].grade){
122                 cal=i+1;
123         }
124         if(some[ep[i].id].rank>cal){
125            some[ep[i].id].rank=cal;
126            some[ep[i].id].r="E";
127         }
128     }
129 
130 
131     map<string,somenode>::iterator it;
132     for(i=0;i<m;i++){
133         cin>>num;
134         it=some.find(num);
135         if(it==some.end()){
136             cout<<"N/A"<<endl;
137         }
138         else{
139             cout<<some[num].rank<<" "<<some[num].r<<endl;
140         }
141     }
142 
143     delete []ap;
144     delete []cp;
145     delete []mp;
146     delete []ep;
147     return 0;
148 }

 

posted @ 2015-08-04 21:00  Deribs4  阅读(220)  评论(0)  编辑  收藏  举报