2014ACM/ICPC亚洲区广州站 && HDU Song Jiang's rank list （排序）

Song Jiang's rank list

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 805 Accepted Submission(s): 410


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


Source
2014ACM/ICPC亚洲区广州站-重现赛（感谢华工和北大）


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解析：直接按题意排序模拟即可

AC代码：

#include 
using namespace std;

struct Node{
    string s;
    int k;
    int r;
    int rr;
} ;
Node a[202];

bool cmp(Node u, Node v){
    if(u.k == v.k) return u.s < v.s;
    return u.k > v.k;
}

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif // sxk
    int n, m;
    string str;
    int k;
    while(scanf("%d", &n) == 1 && n){
        for(int i=0; i>str>>k;
            a[i] = Node{str, k, 0, 0};
        }
        sort(a, a+n, cmp);
        for(int i=0; i>str;
            for(int j=0; j