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

 

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 # include <algorithm>
 5 # include <string>
 6 # include <map>
 7 # include <cmath>
 8 # include <queue>
 9 # include <list>
10 # define LL long long
11 using namespace std ;
12 
13 struct ren
14 {
15     char name[100] ;
16     int num;
17 }a[300];
18 
19 bool cmp(ren a,ren b)
20 {
21     if(a.num!=b.num)
22         return a.num > b.num;
23     return strcmp(a.name,b.name) <0;
24 }
25 
26 map<string,int> mp1 ;
27 map<string,int> mp2 ;
28 
29 int main()
30 {
31     //freopen("in.txt","r",stdin) ;
32     int n , m ;
33     while(scanf("%d" , &n) != EOF)
34     {
35         if (n == 0)
36             break ;
37         mp1.clear() ;
38         mp2.clear() ;
39         int i , j ;
40         char ch[200] ;
41         for (i = 0 ; i < n ; i++)
42             scanf("%s %d" , a[i].name , &a[i].num) ;
43         sort(a , a+n , cmp) ;
44         int sum = 0 ;
45         for (i = 0 ; i < n ; i++)
46         {
47             printf("%s %d\n" , a[i].name , a[i].num) ;
48         }
49 
50         for (i = 0 ; i < n ; i++)
51         {
52             sum=0;
53             for(j = 0 ; j < i ; j++)
54             {
55                 if(a[j].num > a[i].num)
56                     sum++;
57             }
58             mp1[a[i].name]=sum+1;
59         }
60         for (i = 0 ; i < n ; i++)
61         {
62             sum=0;
63             for(j = 0 ; j < i ; j++)
64             {
65                 if(a[i].num==a[j].num)
66                     sum++;
67             }
68             mp2[a[i].name]=sum+1;
69         }
70          scanf("%d" , &m) ;
71          while(m--)
72          {
73              scanf("%s" , ch) ;
74              if (mp2[ch] == 1)
75                 printf("%d\n" , mp1[ch]) ;
76              else
77                 printf("%d %d\n" , mp1[ch] , mp2[ch]) ;
78          }
79     }
80     return 0;
81 }
View Code

 

posted @ 2015-09-30 20:34  __Meng  阅读(216)  评论(0编辑  收藏  举报