hdu 5131 (2014广州现场赛 E题)

题意：对给出的好汉按杀敌数从大到小排序，若相等，按字典序排。M个询问，询问名字输出对应的主排名和次排名。（排序之后）主排名是在该名字前比他杀敌数多的人的个数加1，次排名是该名字前和他杀敌数相等的人的个数加1，（也就是杀敌数相等，但是字典序比他小的人数加1）

Sample Input
5
WuSong 12
LuZhishen 12
SongJiang 13
LuJunyi 1
HuaRong 15
5 //m
WuSong
LuJunyi
LuZhishen
HuaRong
SongJiang
0

Sample Output
HuaRong 15
SongJiang 13
LuZhishen 12
WuSong 12
LuJunyi 1
3 2//主排名 次排名
5
3
1
2

 # include <iostream>
 # include <cstdio>
 # include <cstring>
 # include <algorithm>
 # include <string>
 # include <map>
 # include <cmath>
 # include <queue>
 # include <list>
 # define LL long long
 using namespace std ;

 struct ren
 {
     char name[] ;
     int num;
 }a[];

 bool cmp(ren a,ren b)
 {
     if(a.num!=b.num)
         return a.num > b.num;
     return strcmp(a.name,b.name) <;
 }

 map<string,int> mp1 ;
 map<string,int> mp2 ;

 int main()
 {
     //freopen("in.txt","r",stdin) ;
     int n , m ;
     while(scanf("%d" , &n) != EOF)
     {
         if (n == )
             break ;
         mp1.clear() ;
         mp2.clear() ;
         int i , j ;
         char ch[] ;
         for (i =  ; i < n ; i++)
             scanf("%s %d" , a[i].name , &a[i].num) ;
         sort(a , a+n , cmp) ;
         int sum =  ;
         for (i =  ; i < n ; i++)
         {
             printf("%s %d\n" , a[i].name , a[i].num) ;
         }

         for (i =  ; i < n ; i++)
         {
             sum=;
             for(j =  ; j < i ; j++)
             {
                 if(a[j].num > a[i].num)
                     sum++;
             }
             mp1[a[i].name]=sum+;
         }
         for (i =  ; i < n ; i++)
         {
             sum=;
             for(j =  ; j < i ; j++)
             {
                 if(a[i].num==a[j].num)
                     sum++;
             }
             mp2[a[i].name]=sum+;
         }
          scanf("%d" , &m) ;
          while(m--)
          {
              scanf("%s" , ch) ;
              if (mp2[ch] == )
                 printf("%d\n" , mp1[ch]) ;
              else
                 printf("%d %d\n" , mp1[ch] , mp2[ch]) ;
          }
     }
     return ;
 }