hdu 5131 Song Jiang's rank list 【2014ACM/ICPC亚洲区广州站-重现赛】

Song Jiang's rank list

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 966    Accepted Submission(s): 502

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
Then N lines follow. Each line contains a string S and an integer K(0
The next line is an integer M (0
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

 

题目大意：

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）。

解题思路：

这个题其实就是一个结构体排序，有点水啊。。。。

直接上代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define MM(a) memset(a,0,sizeof(a))

typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 300+5;
const int mod = 1000000007;
const double eps = 1e0-7;
struct node
{
    char str[maxn];
    int peo;
} arr[maxn];
bool cmp(node a, node b)
{
    if(a.peo != b.peo)
        return a.peo > b.peo;
    return strcmp(a.str, b.str) < 0;
}
char st[maxn];
int main()
{
    int n, m;
    while(cin>>n, n)
    {
        for(int i=0; i>arr[i].str>>arr[i].peo;
        sort(arr, arr+n, cmp);

        for(int i=0; i>m;
        while(m--)
        {
            cin>>st;
            for(int i=0; i=0; j--)
                        if(arr[k].peo == arr[j].peo)
                            t++;
                    if(t == 1)
                        cout<