1012. The Best Rank

1012. The Best Rank (25)

时间限制
400 ms
内存限制
32000 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 Algrbra), 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
  1 #include <iostream>
2 #include <fstream>
3 #include <vector>
4 #include <string>
5 #include <algorithm>
6 #include <map>
7 #include <stack>
8 #include <cmath>
9 #include <queue>
10 #include <set>
11 #include <list>
12 #include <stdlib.h>
13 #include <string.h>
14
15
16 using namespace std;
17
18 class Student
19 {
20 public:
21 string id;
22 double M;
23 double C;
24 double E;
25 double A;
26 int rank_a;
27 int rank_c;
28 int rank_m;
29 int rank_e;
30 int best_rank;
31 char best;
32
33 };
34
35 bool compA( const Student &stu1 , const Student &stu2 )
36 {
37 return stu1.A > stu2.A;
38 }
39
40 bool compC( const Student &stu1 , const Student &stu2 )
41 {
42 return stu1.C > stu2.C;
43 }
44
45 bool compM( const Student &stu1 , const Student &stu2 )
46 {
47 return stu1.M > stu2.M;
48 }
49
50 bool compE( const Student &stu1 , const Student &stu2 )
51 {
52 return stu1.E > stu2.E;
53 }
54
55
56
57 int main()
58 {
59
60
61 int N,M;
62
63 vector<Student> all;
64
65
66 cin >> N >> M;
67
68 for( int i = 0 ; i < N ; ++i )
69 {
70 Student stu;
71 cin >> stu.id;
72 cin >> stu.C >> stu.M >> stu.E;
73
74 stu.A = (stu.C + stu.M + stu.E)/3;
75 all.push_back(stu);
76 }
77
78 sort( all.begin() , all.end() , compA );
79
80 for( int i = 0 ; i < all.size() ; ++i )
81 {
82 if( i == 0 )
83 {
84 all[i].rank_a = i+1;
85 }
86 else
87 {
88 if( all[i - 1].A == all[i].A )
89 {
90 all[i].rank_a = all[i-1].rank_a;
91 }
92 else
93 {
94 all[i].rank_a = i+1;
95 }
96 }
97 }
98
99 sort( all.begin() , all.end() , compC );
100
101 for( int i = 0 ; i < all.size() ; ++i )
102 {
103 if( i == 0 )
104 {
105 all[i].rank_c = i+1;
106 }
107 else
108 {
109 if( all[i - 1].C == all[i].C )
110 {
111 all[i].rank_c = all[i-1].rank_c;
112 }
113 else
114 {
115 all[i].rank_c = i+1;
116 }
117 }
118 }
119
120
121 sort( all.begin() , all.end() , compM );
122
123 for( int i = 0 ; i < all.size() ; ++i )
124 {
125 if( i == 0 )
126 {
127 all[i].rank_m = i+1;
128 }
129 else
130 {
131 if( all[i - 1].M == all[i].M )
132 {
133 all[i].rank_m = all[i-1].rank_m;
134 }
135 else
136 {
137 all[i].rank_m = i+1;
138 }
139 }
140 }
141
142 sort( all.begin() , all.end() , compE );
143
144 for( int i = 0 ; i < all.size() ; ++i )
145 {
146 if( i == 0 )
147 {
148 all[i].rank_e = i+1;
149 }
150 else
151 {
152 if( all[i - 1].E == all[i].E )
153 {
154 all[i].rank_e = all[i-1].rank_e;
155 }
156 else
157 {
158 all[i].rank_e = i+1;
159 }
160 }
161 }
162
163 for( int i = 0 ; i < all.size() ; ++i )
164 {
165 int a = all[i].rank_a;
166 int c = all[i].rank_c;
167 int m = all[i].rank_m;
168 int e = all[i].rank_e;
169
170 if( a <= c && a <=m && a <= e )
171 {
172 all[i].best_rank = a;
173 all[i].best = 'A';
174 }
175 else if( c <= a && c <=m && c <= e )
176 {
177 all[i].best_rank = c;
178 all[i].best = 'C';
179 }
180 else if( m <= a && m <=c && m <= e )
181 {
182 all[i].best_rank = m;
183 all[i].best = 'M';
184 }
185 else
186 {
187 all[i].best_rank = e;
188 all[i].best = 'E';
189 }
190 }
191
192 for( int i = 0 ; i < M ; ++i )
193 {
194 string id;
195 cin >> id;
196
197 int p = 0;
198 for( ; p< all.size() ; ++p )
199 {
200 if( all[p].id == id )
201 {
202 break;
203 }
204 }
205
206 if( p == all.size() )
207 {
208 cout << "N/A" << endl;
209 }
210 else
211 {
212 cout << all[p].best_rank << " " << all[p].best << endl;
213 }
214 }
215
216
217 return 0;
218 }


posted on 2012-01-30 13:00  坑王  阅读(245)  评论(0编辑  收藏  举报

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