Rank List
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Time Limit: 10000MS |
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Memory Limit: 65536K |
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Total Submissions: 6561 |
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Accepted: 2091 |
Description
Li Ming is a good student. He always asks the teacher about his rank in his class after every exam, which makes the teacher very tired. So the teacher gives him the scores of all the student in his class and asked him to get his rank by himself. However, he has so many classmates, and he can’t know his rank easily. So he tends to you for help, can you help him?
Input
The first line of the input contains an integer N (1 <= N <= 10000), which represents the number of student in Li Ming’s class. Then come N lines. Each line contains a name, which has no more than 30 letters. These names represent all the students in Li Ming’s class and you can assume that the names are different from each other.
In (N+2)-th line, you'll get an integer M (1 <= M <= 50), which represents the number of exams. The following M parts each represent an exam. Each exam has N lines. In each line, there is a positive integer S, which is no more then 100, and a name P, which must occur in the name list described above. It means that in this exam student P gains S scores. It’s confirmed that all the names in the name list will appear in an exam.
Output
The output contains M lines. In the i-th line, you should give the rank of Li Ming after the i-th exam. The rank is decided by the total scores. If Li Ming has the same score with others, he will always in front of others in the rank list.
Sample Input
3
Li Ming
A
B
2
49 Li Ming
49 A
48 B
80 A
85 B
83 Li Ming
Sample Output
1
2
Source
POJ Monthly,Li Haoyuan
給定每個(gè)人的成績，查詢某一人的名次。
用MAP建立人名和成績的對應(yīng)關(guān)系，用cnt數(shù)組（最多5000個(gè)元素）記錄成績?yōu)槟硞€(gè)分?jǐn)?shù)的人數(shù)，不過由于總?cè)藬?shù)較少(最多只有10000人),直接遍歷也不比建立計(jì)數(shù)排序數(shù)組多用多少時(shí)間,計(jì)數(shù)排序的優(yōu)勢并不顯著.
用hash函數(shù)或者二分查找也應(yīng)該能解決這個(gè)問題.
/**//*Source?Code
Problem:?2153??User:?y09
Memory:?1236K??Time:?1204MS?
Language:?C++??Result:?Accepted?
Source?Code?*/
#include?<iostream>
#include<string>
#include<map>
using?namespace?std;
int?main(int?argc,?char?*argv[])
{
????int?n,m;
????int?i,j;
????char?str[200];
????string?str1;
????
????map<string?,int>score;
????
????scanf("%d",&n);
????getchar();
????for?(i=0;i<n;i++?)
????{
????????gets(str);
????????str1=str;
????????score[str1]=0;
????}
????
????int?cnt[5005]={0};
????
????scanf("%d",&m);
????string?li="Li?Ming";
????int?rank=0;
????int?s=0;
????int?temp=0;
????int?temp2=0;
????int?num;
????for(int?k=0;k<m;k++)
????{
????????for(i=0;i<n;i++)
????????{
????????????scanf("%d",&num);
????????????getchar();
????????????gets(str);
????????????str1=str;
????????????temp2=score[str1];
????????????score[str1]=num+temp2;
????????????cnt[num+temp2]++;
????????}
????????s=score[li];
????????rank=1;
????????temp+=100;
????????for(i=temp;i>s;i--)
????????{
????????????rank+=cnt[i];
????????????cnt[i]=0;
????????}
????????for(i=s;i>=0;i--)
????????????cnt[i]=0;
????????printf("%d\n",rank);
????}
????return?0;
}