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:
2
5
1234567890001 95
1234567890005 100
1234567890003 95
1234567890002 77
1234567890004 85
4
1234567890013 65
1234567890011 25
1234567890014 100
1234567890012 85
Sample Output:
9
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