The Best Rank (25)（排名算法）

 

 

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

 

排名不应该只是简单的排序算法：

比如说  100，100，99，……   得99分是人是 第二名，而不是第三名。

所以，应该先对分数进行排序，然后在更新名次，算法如下：

        

    

 1      SS[0].ra=1； //第一个肯定是第一名

 2 

 3          int count=1;

 4 

 5          for(i=1;i<n;i++)

 6 

 7             {

 8 

 9                   if(SS[i].A==SS[i-1].A)//A为平均分

10 

11                   {

12 

13                       SS[i].ra=SS[i-1].ra;//ra 平均分排名

14 

15                   count++;//如果分数相同，那么名次和前一个相同，计数器++

16 

17                   }

18 

19                 else

20 

21                     {

22 

23                        SS[i].ra=SS[i-1].ra+count;

24 

25                         count=1;//计算器记得还原

26 

27                     }

28 

29          }

 

 

 

AC代码：

 

  1 #include <iostream>

  2 

  3 #include <algorithm>

  4 

  5 #include <string>

  6 

  7 using namespace std;

  8 

  9  

 10 

 11 struct stu

 12 

 13 {

 14 

 15    string ID;

 16 

 17    int A,C,M,E;

 18 

 19    int ra,rc,rm,re;

 20 

 21 };

 22 

 23  

 24 

 25 stu SS[10001];

 26 

 27  

 28 

 29  

 30 

 31  

 32 

 33 bool cmp1(stu a,stu b)

 34 

 35 {

 36 

 37     return a.A>b.A;

 38 

 39 }

 40 

 41  

 42 

 43  

 44 

 45 bool cmp2(stu a,stu b)

 46 

 47 {

 48 

 49     return a.C>b.C;

 50 

 51 }

 52 

 53  

 54 

 55 bool cmp3(stu a,stu b)

 56 

 57 {

 58 

 59     return a.M>b.M;

 60 

 61 }

 62 

 63  

 64 

 65  

 66 

 67 bool cmp4(stu a,stu b)

 68 

 69 {

 70 

 71     return a.E>b.E;

 72 

 73 }

 74 

 75  

 76 

 77 int main()

 78 

 79 {

 80 

 81  

 82 

 83       int n;

 84 

 85       while(cin>>n)

 86 

 87       {

 88 

 89             int m,i;

 90 

 91          cin>>m;

 92 

 93      

 94 

 95          for(i=0;i<n;i++)

 96 

 97          {

 98 

 99        

100 

101                cin>>SS[i].ID>>SS[i].C>>SS[i].M>>SS[i].E;

102 

103          }

104 

105  

106 

107  

108 

109           for(i=0;i<n;i++)

110 

111          {

112 

113                 SS[i].A=(SS[i].C+SS[i].M+SS[i].E)/3;

114 

115                     SS[i].ra=1;

116 

117                     SS[i].rc=1;

118 

119                     SS[i].rm=1;

120 

121                     SS[i].re=1;

122 

123          }

124 

125        

126 

127  

128 

129            

130 

131        sort(SS,SS+n,cmp1);

132 

133          int count=1;

134 

135          for(i=1;i<n;i++)

136 

137             {

138 

139                   if(SS[i].A==SS[i-1].A)

140 

141                   {

142 

143                       SS[i].ra=SS[i-1].ra;

144 

145                               count++;

146 

147                   }

148 

149                 else

150 

151                     {

152 

153                        SS[i].ra=SS[i-1].ra+count;

154 

155                         count=1;

156 

157                     }

158 

159          }

160 

161  

162 

163  

164 

165             sort(SS,SS+n,cmp2);

166 

167  

168 

169           count=1;

170 

171          for(i=1;i<n;i++)

172 

173          {

174 

175              if(SS[i].C==SS[i-1].C)

176 

177                   {

178 

179                       SS[i].rc=SS[i-1].rc;

180 

181                               count++;

182 

183                   }

184 

185                 else

186 

187                     {

188 

189                        SS[i].rc=SS[i-1].rc+count;

190 

191                         count=1;

192 

193                     }

194 

195          }

196 

197  

198 

199  

200 

201             sort(SS,SS+n,cmp3);

202 

203          count=1;

204 

205          for(i=1;i<n;i++)

206 

207          {

208 

209              if(SS[i].M==SS[i-1].M)

210 

211                   {

212 

213                       SS[i].rm=SS[i-1].rm;

214 

215                               count++;

216 

217                   }

218 

219                 else

220 

221                     {

222 

223                        SS[i].rm=SS[i-1].rm+count;

224 

225                         count=1;

226 

227                     }

228 

229          }

230 

231  

232 

233  

234 

235             sort(SS,SS+n,cmp4);

236 

237          count=1;

238 

239          for(i=1;i<n;i++)

240 

241          {

242 

243              if(SS[i].E==SS[i-1].E)

244 

245                   {

246 

247                       SS[i].re=SS[i-1].re;

248 

249                               count++;

250 

251                   }

252 

253                 else

254 

255                     {

256 

257                        SS[i].re=SS[i-1].re+count;

258 

259                         count=1;

260 

261                     }

262 

263          }

264 

265        

266 

267  

268 

269               for(i=0;i<m;i++)

270 

271          {

272 

273                   string goal;int j;

274 

275               cin>>goal;

276 

277                   for(j=0;j<n;j++)

278 

279                         if(SS[j].ID==goal) break;

280 

281  

282 

283           if(j==n) cout<<"N/A"<<endl;

284 

285             else{           

286 

287                   char high='E';

288 

289                   int temp=SS[j].re;

290 

291                   if(temp>=SS[j].rm)

292 

293                   {

294 

295                      high='M';

296 

297                      temp=SS[j].rm;

298 

299                   }

300 

301                   if(temp>=SS[j].rc)

302 

303                   {

304 

305                      high='C';

306 

307                      temp=SS[j].rc;

308 

309                   }

310 

311                   if(temp>=SS[j].ra)

312 

313                   {

314 

315                      high='A';

316 

317                      temp=SS[j].ra;

318 

319                   }

320 

321  

322 

323                   cout<<temp<<" "<<high<<endl;

324 

325             }

326 

327          }

328 

329       }

330 

331   return 0;

332 

333 }

334 

335  

View Code