2013
11-10

# Rank List

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?

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.

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.

3
Li Ming
A
B
2
49 Li Ming
49 A
48 B
80 A
85 B
83 Li Ming


1
2


//* @author: 82638882@163.com
import java.util.HashMap;
import java.util.Scanner;

public class Main
{
public static void main(String[] args)
{
Scanner in=new Scanner(System.in);
int n=in.nextInt();
int[] arr=new int[n];
HashMap< String,Integer> hs=new HashMap< String,Integer>();
String s=in.nextLine();
int tag=-1;
for(int i=0;i< n;i++)
{
s=in.nextLine();
s=s.trim();
if(s.equals("Li Ming"))tag=i;
hs.put(s, i);
}
int m=in.nextInt();
for(int i=0;i< m;i++)
{
for(int j=0;j< n;j++)
{
int num=in.nextInt();
s=in.nextLine();
s=s.trim();
int u=hs.get(s);
arr[u]+=num;
}
int ww=arr[tag];
int cnt=1;
for(int j=0;j< n;j++)
if(arr[j]>arr[tag])cnt++;
System.out.println(cnt);
}

}

}

1. “再把所有不和该节点相邻的节点着相同的颜色”，程序中没有进行不和该节点相邻的其他节点是否相邻进行判断。再说求出来的也不一样是颜色数最少的

2. Gucci New Fall Arrivals

This is really nice to know. I hope it will be successful in the future. Good job on this and keep up the good work.