结构体+sort方法

昨天做了一道简单但很麻烦的题,我只能想到结构体,并用了STL的sort方法解决了它.不过从中有许多细节问题.

题目:

 

Problem Description

  Lcy wanted to choose 50 ACMers from m players to join HDU-ACM team.
He made n competitions , and now is your task to make the ranklist.
Here’re some instructions :
Effective score : the sum of the best n-2 competitions’ score.
One’s score in one competition : number of problems he(or she) solved divided the sum of problems all players solved.

The ranklist is made , of course , by the effective score , what if some players have the same score ?Follow these rules : girls always come first , same again and younger first , again ? In lexicographic order.

Input

  For each case , two integers in the first line : n (3<= n <= 10)and m(m<=1000)
The next m lines are someone’s information :
name , sex (F，M), grade and n integer numbers which means one’s number of solved problem in the ith competition.

Output

Output the first fifty players in order . Output all if m<50.

Sample Input

3 2
WXL F 08 3 4 5
HH M 08 3 4 5

Sample Output

WXL
HH

 

代码如下:

 

#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
bool dic(char *a, char *b)
{
    int p;
    p=strcmp(a,b);
    if(p>0)
        return false;
    else return true;
}
typedef struct
{
    char name[21];
    char sex;
    int grade;
    float score;
    int s[11];
}node;
node a[1001];

bool comp(node a, node b)
{
    if(a.score != b.score)
        return a.score>b.score;
    else
    {
        if(a.sex != b.sex)
            return a.sex<b.sex;
        else if(a.grade != b.grade)
            return a.grade>b.grade;
        else
            return dic(a.name,b.name);
    }
}

int main()
{
    int n,m,i,j,k;
    float b[11],p,f[11];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0; i<m; i++)
        {
            getchar();
            scanf("%s %c %d",&a[i].name,&a[i].sex,&a[i].grade);
            for(j=0; j<n; j++)
            {
                scanf("%d",&a[i].s[j]);
            }
        }

        for(i=0; i<n; i++)
        {
            b[i]=0;
            for(j=0; j<m; j++)
            {
                b[i]+=a[j].s[i];
            }
        }
        for(i=0; i<m; i++)
        {
            a[i].score=0;
            for(j=0; j<n; j++)
            {
                p=(float)a[i].s[j]/b[j];
                f[j]=p;
            }
            sort(f,f+n);
            for(k=n-1; k>1; k--)
            {
                a[i].score+=f[k];
            }
        }
        sort(a,a+m,comp);
        for(i=0; i<m; i++)
        {
            cout<<a[i].name<<endl;
            if(i==49)break;
        }
    }
    return 0;
}

来源：https://www.cnblogs.com/chenjunyu/archive/2009/05/04/1448920.html