PAT Ranking (排名)

Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate the final rank.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive number N (<=100), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (<=300), the number of testees, and then K lines containing the registration number (a 13-digit number) and the total score of each testee. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in one line the total number of testees. Then print the final ranklist in the following format:

registration_number final_rank location_number local_rank

The locations are numbered from 1 to N. The output must be sorted in nondecreasing order of the final ranks. The testees with the same score must have the same rank, and the output must be sorted in nondecreasing order of their registration numbers.

Sample Input:

1234567890001 95

1234567890005 100

1234567890003 95

1234567890002 77

1234567890004 85

1234567890013 65

1234567890011 25

1234567890014 100

1234567890012 85

Sample Output:

1234567890005 1 1 1

1234567890014 1 2 1

1234567890001 3 1 2

1234567890003 3 1 2

1234567890004 5 1 4

1234567890012 5 2 2

1234567890002 7 1 5

1234567890013 8 2 3

1234567890011 9 2 4

先分组排名，再全部排名

  1 #include <iostream>
  2 
  3 #include <string>
  4 
  5 #include <algorithm>
  6 
  7 #include <vector>
  8 
  9 using namespace std;
 10 
 11  
 12 
 13 struct stu
 14 
 15 {
 16 
 17    int loc,frank,lrank,gra;
 18 
 19    string name;
 20 
 21 };
 22 
 23  
 24 
 25  
 26 
 27 bool cmp(stu s1,stu s2)
 28 
 29 {
 30 
 31     if(s1.frank==s2.frank)
 32 
 33     return s1.name<s2.name;
 34 
 35  
 36 
 37       return s1.frank<s2.frank;
 38 
 39 }
 40 
 41  
 42 
 43 bool cmpg(stu s1,stu s2)
 44 
 45 {
 46 
 47  
 48 
 49       return s1.gra>s2.gra;
 50 
 51 }
 52 
 53  
 54 
 55 int main()
 56 
 57 {
 58 
 59  
 60 
 61   int n;int k;int rank,count;
 62 
 63   int i,j;
 64 
 65     while(cin>>n)
 66 
 67   {
 68 
 69       vector<stu> total;
 70 
 71      for(i=1;i<=n;i++)
 72 
 73      {
 74 
 75          cin>>k;
 76 
 77             vector<stu> s(k);
 78 
 79         for(j=0;j<k;j++)
 80 
 81             {
 82 
 83                cin>>s[j].name>>s[j].gra;
 84 
 85                s[j].loc=i;
 86 
 87             }
 88 
 89             sort(s.begin(),s.end(),cmpg);
 90 
 91             count=1;
 92 
 93             s[0].lrank=1;
 94 
 95             total.push_back(s[0]);
 96 
 97             for(j=0;j<k-1;j++)
 98 
 99             {
100 
101                   if(s[j].gra==s[j+1].gra)
102 
103                   {
104 
105                         s[j+1].lrank=s[j].lrank;
106 
107                         count++;
108 
109                   }
110 
111                   else
112 
113                   {
114 
115                       s[j+1].lrank=s[j].lrank+count;
116 
117                         count=1;
118 
119                   }
120 
121                   total.push_back(s[j+1]);
122 
123             }
124 
125      }
126 
127  
128 
129           sort(total.begin(),total.end(),cmpg);
130 
131           count=1;
132 
133             total[0].frank=1;
134 
135             for(j=0;j<total.size()-1;j++)
136 
137             {
138 
139                   if(total[j].gra==total[j+1].gra)
140 
141                   {
142 
143                         total[j+1].frank=total[j].frank;
144 
145                         count++;
146 
147                   }
148 
149                   else
150 
151                   {
152 
153                       total[j+1].frank=total[j].frank+count;
154 
155                         count=1;
156 
157                   }
158 
159             }
160 
161  
162 
163             sort(total.begin(),total.end(),cmp);
164 
165             cout<<total.size()<<endl;
166 
167             for(i=0;i<total.size();i++)
168 
169                   cout<<total[i].name<<" "<<total[i].frank<<" "<<total[i].loc<<" "<<total[i].lrank<<endl;
170 
171    
172 
173   }
174 
175   return 0;
176 
177 }
178 
179

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