PAT 1025. PAT Ranking (25) 数据结构，排序

1025. PAT Ranking (25)

时间限制200 ms内存限制65536 kB代码长度限制16000 B判题程序Standard作者CHEN, YueProgramming 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 yourjob 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 thenK 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_rankThe 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

和前面一道一模一样。。就是先按考场排，定下小考场rank。在按分数排，定下总rank

#include <cstdio>#include<string.h>#include<string>#include<map>#include<vector>#include<stdio.h>#include<set>#include<iostream>#include <algorithm>using namespace std;struct ZZ{string id;int score;int xrank;int drank;int kaochang;};bool cmp1 (ZZ a,ZZ b){if(a.kaochang!=b.kaochang) return a.kaochang<b.kaochang;return a.score>b.score;}bool cmp2 (ZZ a,ZZ b){if(a.score!=b.score)return a.score>b.score;return a.id<b.id;}vector<ZZ> Q;int main(){int k;cin>>k;int sum=0;for(int i=1;i<=k;i++){int n;cin>>n;sum+=n;for(int j=0;j<n;j++){ZZ tmp;cin>>tmp.id>>tmp.score;tmp.kaochang=i;Q.push_back(tmp);}}cout<<sum<<endl;sort(Q.begin(),Q.end(),cmp1);int p=0;int head=0;int tail=0;while(p<Q.size()){if(Q[p].kaochang==Q[head].kaochang) {p++;continue;}tail=p-1;Q[head].xrank=1;for(int i=head+1;i<=tail;i++){if(Q[i].score!=Q[i-1].score) Q[i].xrank=i-head+1;else Q[i].xrank=Q[i-1].xrank;}p++;head=tail+1;}tail=p-1;Q[head].xrank=1;for(int i=head+1;i<=tail;i++){if(Q[i].score!=Q[i-1].score) Q[i].xrank=i-head+1;else Q[i].xrank=Q[i-1].xrank;}sort(Q.begin(),Q.end(),cmp2);Q[0].drank=1;for(int i=1;i<Q.size();i++){if(Q[i].score!=Q[i-1].score) Q[i].drank=i+1;else Q[i].drank=Q[i-1].drank;}for(int i=0;i<Q.size();i++)cout<<Q[i].id<<' '<<Q[i].drank<<' '<<Q[i].kaochang<<' '<<Q[i].xrank<<endl;return 0;}