H-index因子

Problem Description

Paper quality and quantity have long been used to measure a research's scientific productivity and scientific impact. Citation, which is the total times a paper has been cited, is a common means to judge importance of a paper. However, with all these factors varying, a collegiate committee has problems when judging which research is doing better. For this reason, H-index is proposed and now widely used to combine the above factors and give accurate judgement. H-index is defined as:
A scientist has index h if h of [his] Np papers have at least h citations each, and the other(Np-h) papers have at most h citations each.
In other words, a scholar with an index of h has published h papers each of which has been cited by others at least h times. Note that H-index is always an integer. It's said that achiveing H-index of 18 means one is fully quality to be a professor, and H-index of 45 or higher could mean membership in the United States Academy of Sciences.
You are to calculate the H-index for all the researchers in the list, base on the given information.

Input

There are multiple scenarios in the input, terminated by a single zero(0).
Each of the scenarios begin with an integer N(1<=N<=100), means that there are N papers. N lines follow, each contain a string(not exceeding 20 characters long), representing the author of the corresponding paper, without white spaces in-between. Though it would be common for one paper written by several authors, there would be exactly one author of each of these papers. Finally, there are N lines of strings, containing '0's and '1's. If the j-th character in the i-th line is '1', it means the i-th paper cites the j-th paper. A paper could never cite itself.

Output

For each scenario, output as many lines as the number of authors given. Each line contains the author's name and his H-index. The list should be sorted first by H-index in descending order, than by name in alphabetic order(Actually, ASCII order. So 'B' is prior to 'a').
Output a blank line after each scenario.

Sample Input

4
Peter
Peter
Bob
Bob
0000
1000
1100
0100
0

Sample Output

Peter 2
Bob 0
/*
思路：引用，不能引用文章自己本身，但可以引用同一个作者的其他文章（引用的是1，没有引用是0）
统计引用次数后，对同一个作者的引用次数排降序，再编号(把同一个人名放在一起)
比较编号<引用次数，H指数加一次
再对H指数排降序，对作者名字排升序
*/
#include<cstdio>
#include<string.h>
#include<iostream>
using namespace std;
struct msc//存放原来的
{
    char s[21];//作者
    char b[104];//论文
    int flot;//存引用次数
};
struct mmx//存放的是统计出每个作者，和h指数
{
   char str[21];//作者
   int sum;//h指数
};
int main()
{
    msc a[104];
    int n,j,i,m,num,k,w,h;
    char t[21];
     while(scanf("%d",&n)!=EOF&&n)
     {
         for(i=1;i<=n;i++)
          {
              cin>>a[i].s;
            a[i].flot=0;
          }
            for(i=1;i<=n;i++)
               for(j=1;j<=n;j++)
                cin>>a[i].b[j];

            for(i=1;i<=n;i++)//统计引用次数
              for(j=1;j<=n;j++)
                if(a[i].b[j]=='1'&&(i!=j))
                    a[j].flot++;

              for(i=1;i<n;i++)//排序，按作者名排序，再按引用次数排序，降序
              {
                for(j=i+1;j<=n;j++)
                {
                if(strcmp(a[i].s,a[j].s)<0)
                    {
                        strcpy(t,a[i].s);strcpy(a[i].s,a[j].s);strcpy(a[j].s,t);
                        w=a[i].flot;a[i].flot=a[j].flot;a[j].flot=w;
                    }
                else if(strcmp(a[i].s,a[j].s)==0)
					{
                        if(a[i].flot<a[j].flot)
                            {w=a[i].flot;a[i].flot=a[j].flot;a[j].flot=w;
                            strcpy(t,a[i].s);strcpy(a[i].s,a[j].s);strcpy(a[j].s,t);}
					}
				}
			  }

				mmx x[104]; k=1;
                       for(i=1;i<=n;i=j)//统计h指数：把每位作者的论文的引用次数排序之后，从1开始编号，然后引用次数和编号比较
                       {      m=1;num=0;

                           for(j=i;strcmp(a[i].s,a[j].s)==0;j++)
                           {
                               if(m<=a[j].flot){m++,num++;}//编号<次数。。h指数加一次
                           }
                               strcpy(x[k].str,a[i].s);
							   x[k].sum=num;
							   k++;
                      }
                       h=k-1;
                for(i=1;i<h;i++)//排序，按h指数排序，降序，再按作者名排序，排升序
					{
                        for(j=i+1;j<=h;j++)

                       {
                            if(x[i].sum<x[j].sum)
                            {
			                     strcpy(t,x[i].str);strcpy(x[i].str,x[j].str);strcpy(x[j].str,t);
                                 w=x[i].sum; x[i].sum=x[j].sum;x[j].sum=w;
                            }
			               else if(x[i].sum==x[j].sum)
                            {
                               if(strcmp(x[i].str,x[j].str)>0)
                               {
                                strcpy(t,x[i].str);strcpy(x[i].str,x[j].str);strcpy(x[j].str,t);
                                w=x[i].sum; x[i].sum=x[j].sum;x[j].sum=w;
                                }
                            }
			          }
		           }
			  for(i=1;i<k;i++)//输出
                          printf("%s %d\n",x[i].str,x[i].sum);
		         cout<<endl;
}
return 0;
}