HDU 5131 Song Jiang's rank list （确定排名经典问题）

《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

5WuSong 12LuZhishen 12SongJiang 13LuJunyi 1HuaRong 155WuSongLuJunyiLuZhishenHuaRongSongJiang0

Sample Output

HuaRong 15SongJiang 13LuZhishen 12WuSong 12LuJunyi 13 25312

题意：

水浒传里的排名按杀敌数排（exm???）,杀敌数相同的谁名字字典序靠前谁就靠前。

思路：

排名问题的经典问题。定义结构体使用强大的sort中的cmp就ok了。

注意其中排名一样的，这个排名里的第一个只输出排名，第二个输出排名  2 ，第三个输出排名  3

存在并列第三，如果两人并列，下一人的名次是第五。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <algorithm>#include <math.h>#include <vector>#include <iostream>#include <cstdio>using namespace std;struct aa{    string x;    int y;    int mc;    int mc2;}a[500];struct bb{    int a;    int b;}ans[500];bool cmp(aa p, aa q){    if(p.y==q.y)        return p.x<q.x;    return p.y>q.y;}int main(){    int n;    while(scanf("%d",&n)&&n)    {for(int i=0;i<n;i++)        cin>>a[i].x>>a[i].y;    sort(a,a+n,cmp);    int mc=1,mc2=1,maxfx=a[0].y;    for(int i=0;i<n;i++)    {        if(a[i].y!=maxfx)        {mc+=(mc2-1);mc2=1;///这里细节注意一下，对于每个新的杀敌数，mc2更新到1，mc更新mc2的个数，因为mc2后自加，故-1        maxfx=a[i].y;}        a[i].mc=mc;        a[i].mc2=mc2++;    }    int m;    scanf("%d",&m);    for(int i=0;i<m;i++)    {        string temp;        cin>>temp;        for(int j=0;j<n;j++)        {            if(a[j].x==temp)                {ans[i].a=a[j].mc;                ans[i].b=a[j].mc2;break;                }        }    }    for(int i=0;i<n;i++)        cout<<a[i].x<<" "<<a[i].y<<endl;    for(int i=0;i<m;i++)        if(ans[i].b!=1)        printf("%d %d\n",ans[i].a,ans[i].b);    else printf("%d\n",ans[i].a);    }    return 0;}