HDU 5131 Song Jiang's rank list

Song Jiang's rank list

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

Total Submission(s): 381    Accepted Submission(s): 188

Problem Description

《Shui Hu Zhuan》，also 《Water Margin》was written by Shi Nai'an -- an writer of Yuan and Ming dynasty. 《Shui Hu Zhuan》is one of the Four Great Classical Novels of Chinese literature. It tells a story about 108 outlaws. They came from different backgrounds (including scholars, fishermen, imperial drill instructors etc.), and all of them eventually came to occupy Mout Liang(or Liangshan Marsh) and elected Song Jiang as their leader.

In order to encourage his military officers, Song Jiang always made a rank list after every battle. In the rank list, all 108 outlaws were ranked by the number of enemies he/she killed in the battle. The more enemies one killed, one's rank is higher. If two outlaws killed the same number of enemies, the one whose name is smaller in alphabet order had higher rank. Now please help Song Jiang to make the rank list and answer some queries based on the rank list.

 

Input

There are no more than 20 test cases.

For each test case:

The first line is an integer N (0<N<200), indicating that there are N outlaws.

Then N lines follow. Each line contains a string S and an integer K(0<K<300), meaning an outlaw's name and the number of enemies he/she had killed. A name consists only letters, and its length is between 1 and 50(inclusive). Every name is unique.

The next line is an integer M (0<M<200) ,indicating that there are M queries.

Then M queries follow. Each query is a line containing an outlaw's name.
The input ends with n = 0

 

Output

For each test case, print the rank list first. For this part in the output ,each line contains an outlaw's name and the number of enemies he killed.

Then, for each name in the query of the input, print the outlaw's rank. Each outlaw had a major rank and a minor rank. One's major rank is one plus the number of outlaws who killed more enemies than him/her did.One's minor rank is one plus the number of outlaws who killed the same number of enemies as he/she did but whose name is smaller in alphabet order than his/hers. For each query, if the minor rank is 1, then print the major rank only. Or else Print the major rank, blank , and then the minor rank. It's guaranteed that each query has an answer for it.

 

Sample Input

  
  

   
   5
WuSong 12
LuZhishen 12
SongJiang 13
LuJunyi 1
HuaRong 15
5
WuSong
LuJunyi
LuZhishen
HuaRong
SongJiang
0
  
  

 

Sample Output

  
  

   
   HuaRong 15
SongJiang 13
LuZhishen 12
WuSong 12
LuJunyi 1
3 2
5
3
1
2
  
  

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5131

题目大意：给出n给梁山好汉的名字和杀敌人数，第一排名按杀敌人数排名，杀敌越多，排名越靠前。若杀敌人数相同，按第二排名，即字典序小的名字排在前面。按排名顺序输出名字和杀敌人数。

然后m次询问，每次给出一个好汉的名字，输出其排名，若排名重复（这里杀敌人数相同占同一个名次），还要输出第二排名。

解题思路：排序问题，详解见代码。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char st[205];
struct list
{
    char s[52];
    int k;
}a[305];
int cmb(list a,list b)
{
    if(a.k==b.k)      //第二排名
    {
        if(strcmp(a.s,b.s)<0)
            return 1;
        else
            return 0;
    }
    else
        return a.k>b.k;   //第一排名
}
int main()
{
    int n,i,j,m;
    while(scanf("%d" ,&n)!=EOF&&n)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%s",a[i].s);
            scanf("%d",&a[i].k);
        }
        sort(a+1,a+n+1,cmb);   //排序
        for(i=1;i<=n;i++)
            printf("%s %d\n", a[i].s,a[i].k);
        scanf("%d",&m);
        while(m--)
        {
            int cnt=0;
            scanf("%s",st);
            for(j=1;j<=n;j++)
            {
                if(strcmp(st,a[j].s)==0)       
                {
                    while(j-1>0&&a[j].k==a[j-1].k)   //  如果排在前面的与之杀敌人数相同，则j--找到第一排名名次，cnt代表在第二排名中的名次。
					{
                        cnt++;
						j--;
					}
                    if(cnt)
                        printf("%d %d\n",j,cnt+1);
                    else
                        printf("%d\n", j);
					break;
                }
            }
        }
    }
    return 0;
}