HDU5131 Song Jiang's rank list(模拟)

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题目：

Song Jiang's rank list

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1674 Accepted Submission(s): 943

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

Source

2014ACM/ICPC亚洲区广州站-重现赛（感谢华工和北大）

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Statistic | Submit | Discuss | Note 思路:

梁山的好汉有杀敌数，给了n个人，然后后面跟着它的杀敌数，先输出杀敌数的排行榜（如果杀敌数一样按照字典序排序），然后有q组询问，针对每一个询问，如果没有它的名次上面没有人的杀敌数和他一样，那么就直接输出它的名次，如果有人的杀敌数和他一样，先输出它的名次，然后在输出杀敌数和他一样的人的个数+1.。

直接模拟就好

代码：

#include <cstdio>
#include <cstring>
#include <cctype>
#include <string>
#include <set>
#include <iostream>
#include <stack>
#include <map>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f3f3f3f3f
#define mod 10000007
#define debug() puts("what the fuck!!!")
#define N 500010
#define M 1000000
#define ll long long
using namespace std;
struct node
{string name;int kill_num;int num;
} a[300];
bool cmp(node a,node b)
{if(a.kill_num==b.kill_num)return a.name<b.name;elsereturn a.kill_num>b.kill_num;
}
int main()
{int n,q;while(cin>>n){int len=1;map<string,int>mp;if(n==0)return 0;for(int i=1; i<=n; i++){cin>>a[i].name>>a[i].kill_num;mp[a[i].name]=a[i].kill_num;}sort(a+1,a+n+1,cmp);a[1].num=1;int flag=0;for(int i=1; i<=n; i++){cout<<a[i].name<<" "<<a[i].kill_num<<endl;if(i>1){if(a[i].kill_num==a[i-1].kill_num){a[i].num=a[i-1].num;flag++;}else{a[i].num=a[i-1].num+1+flag;flag=0;}}}cin>>q;string s;while(q--){cin>>s;flag=0;for(int i=1;i<=n;i++){if(mp[s]==a[i].kill_num)flag++;if(a[i].name==s){if(flag>1)printf("%d %d\n",a[i].num,flag);elseprintf("%d\n",a[i].num);break;}}}}return 0;
}

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