PAT 1083 List Grades[简单]

1083 List Grades （25 分）

Given a list of N student records with name, ID and grade. You are supposed to sort the records with respect to the grade in non-increasing order, and output those student records of which the grades are in a given interval.

Input Specification:

Each input file contains one test case. Each case is given in the following format:

N

name[1] ID[1] grade[1]

name[2] ID[2] grade[2]

... ...

name[N] ID[N] grade[N]

grade1 grade2

where name[i] and ID[i] are strings of no more than 10 characters with no space, grade[i] is an integer in [0, 100], grade1 and grade2 are the boundaries of the grade's interval. It is guaranteed that all the grades are distinct.

Output Specification:

For each test case you should output the student records of which the grades are in the given interval [grade1, grade2] and are in non-increasing order. Each student record occupies a line with the student's name and ID, separated by one space. If there is no student's grade in that interval, output NONE instead.

Sample Input 1:

4

Tom CS000001 59

Joe Math990112 89

Mike CS991301 100

Mary EE990830 95

60 100

Sample Output 1:

Mike CS991301

Mary EE990830

Joe Math990112

Sample Input 2:

2

Jean AA980920 60

Ann CS01 80

90 95

Sample Output 2:

NONE

题目大意：给出n个学生，姓名id分数，并给出分数区间，要求在分数区间内的学生按分数排序输出。

//这个就很简单啦，我的AC：

#include <iostream>

#include <algorithm>

#include <vector>

using namespace std;

struct Stu{

    string name,id;

    int sco;

};

vector<Stu> allstu;

vector<Stu> stu;

bool cmp(Stu&a,Stu&b){

    return a.sco>b.sco;

}

int main() {

    int n;

    cin>>n;

    string name,id;

    int sco,low,high;

    for(int i=;i<n;i++){

        cin>>name>>id>>sco;

        allstu.push_back(Stu{name,id,sco});

    }

    cin>>low>>high;

    for(auto it=allstu.begin();it!=allstu.end();){

        if(it->sco<low||it->sco>high){

            it=allstu.erase(it);//

        }else it++;

    }

    if(allstu.size()==){

        cout<<"NONE";

    }else{

        sort(allstu.begin(),allstu.end(),cmp);

        for(int i=;i<allstu.size();i++){

            cout<<allstu[i].name<<" "<<allstu[i].id<<'\n';

        }

    }

    return ;

}

//以下是遇到的问题：

1，关于使用vector::erase函数。

参数是iterator迭代器，转自：https://www.cnblogs.com/zsq1993/p/5930229.html

for(vector<int>::iterator iter=veci.begin(); iter!=veci.end(); iter++)

{

      if( *iter == )

             veci.erase(iter);

}

乍一看这段代码，很正常。其实这里面隐藏着一个很严重的错误：当veci.erase(iter)之后，iter就变成了一个野指针，对一个野指针进行 iter++ 是肯定会出错的。

for(vector<int>::iterator iter=veci.begin(); iter!=veci.end(); iter++)

{

      if( *iter == )

             iter = veci.erase(iter);

}

这段代码也是错误的：1）无法删除两个连续的"3"； 2）当3位于vector最后位置的时候，也会出错（在veci.end()上执行 ++ 操作）

for(vector<int>::iterator iter=veci.begin(); iter!=veci.end(); )

{

     if( *iter == )

          iter = veci.erase(iter);

      else

            iter ++ ;

}

第三种是正确的写法，学习了！！

巴特西