1141. PAT Ranking of Institutions (25)

冷不防 2022-05-28 08:59 205阅读 0赞

After each PAT, the PAT Center will announce the ranking of institutions based on their students' performances. Now you are asked to generate the ranklist.

**Input Specification:**

Each input file contains one test case. For each case, the first line gives a positive integer N (<=105), which is the number of testees. Then N lines follow, each gives the information of a testee in the following format:

ID Score School

where "ID" is a string of 6 characters with the first one representing the test level: "B" stands for the basic level, "A" the advanced level and "T" the top level; "Score" is an integer in \[0, 100\]; and "School" is the institution code which is a string of no more than 6 English letters (case insensitive). Note: it is guaranteed that "ID" is unique for each testee.

**Output Specification:**

For each case, first print in a line the total number of institutions. Then output the ranklist of institutions in nondecreasing order of their ranks in the following format:

Rank School TWS Ns

where "Rank" is the rank (start from 1) of the institution; "School" is the institution code (all in lower case); "TWS" is the **total weighted score** which is defined to be the integer part of "ScoreB/1.5 + ScoreA \+ ScoreT\*1.5", where "ScoreX" is the total score of the testees belong to this institution on level X; and "Ns" is the total number of testees who belong to this institution.

The institutions are ranked according to their TWS. If there is a tie, the institutions are supposed to have the same rank, and they shall be printed in ascending order of Ns. If there is still a tie, they shall be printed in alphabetical order of their codes.

**Sample Input:**

10
    A57908 85 Au
    B57908 54 LanX
    A37487 60 au
    T28374 67 CMU
    T32486 24 hypu
    A66734 92 cmu
    B76378 71 AU
    A47780 45 lanx
    A72809 100 pku
    A03274 45 hypu

**Sample Output:**

5
    1 cmu 192 2
    1 au 192 3
    3 pku 100 1
    4 hypu 81 2
    4 lanx 81 2

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题目大意：

代码：

#include<stdio.h>
    #include<iostream>
    #include<string>
    #include<map>
    #include<algorithm>
    #include<math.h>
    using namespace std;
    struct node
    {
        string firm;
        double sum;
        int sum1;
        int num;
        int Rank;
    }person[100001];
    map<string,struct node> Map;
    bool cmp(struct node a,struct node b)
    {
        if(a.sum1!=b.sum1)
        {
            return a.sum1>b.sum1;
        }
        else if(a.num!=b.num)
        {
            return a.num<b.num;
        }
        else
        {
            return a.firm<b.firm;
        }
    }
    int main()
    {
        int i,j,n,k,t,l,Rank;
        double m;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            string g,f;
            cin>>g>>m>>f;
            for(j=0;j<f.size();j++)
            {
                if(f[j]>='A'&&f[j]<='Z')
                {
                    f[j]=f[j]+32;
                }
            }
            Map[f].firm=f;
            if(g[0]=='A')
            {
                Map[f].sum+=m;
            }
            else if(g[0]=='B')
            {
                Map[f].sum+=(m/1.5);
            }
            else if(g[0]=='T')
            {
                Map[f].sum+=(m*1.5);
            }
            Map[f].num++;
        }
        map<string,struct node>::iterator it;
        l=0;
        for(it=Map.begin();it!=Map.end();it++)
        {
           person[l++]=it->second;
           person[l-1].sum1=(int)person[l-1].sum;
        }
        sort(person,person+l,cmp);
        person[0].Rank=1;
        for(i=1;i<l;i++)
        {
            if(person[i].sum1==person[i-1].sum1)
            {
                person[i].Rank=person[i-1].Rank;
            }
            else
            {
               person[i].Rank=i+1;
            }
        }
        printf("%d\n",l);
        for(i=0;i<l;i++)
        {
            printf("%d %s %d %d\n",person[i].Rank,(person[i].firm).c_str(),person[i].sum1,person[i].num);
        }
        return 0;
    }