PAT倒計時——4天！（排序篇）

還有四天就要考PAT了，這幾天寫博客回顧一下常用的算法，臨時抱一下佛腳哈哈哈~

第一次寫，如有錯誤，請多指教哈~

排序篇

排序是程序員必備技能，也是基本功。不過pat貌似對排序的考察不多，好像就見到插排和歸併，其他的主要穿插在大模擬題中考察，並且可以通過封裝好的sort方法偷懶哈哈哈。

一、選擇和插排

void selectsort(){
    for(int i=0;i<n;i++){
        int index=i;
        for(int j=i+1;j<n;j++){
            if(a[j]<a[index]){
                index=j;
            }
        }
        swap(a[i],a[index]);
    }
}

void insertsort(){

    for(int i=1;i<n;i++){

        int cur=a[i];
        int j=i;
        while(j>=1&&cur<a[j-1]){
            a[j]=a[j-1];
            j--;
        }
        a[j]=cur;
    }

}

二、歸併排序

歸併排序要注意一下，遞歸和非遞歸的排序過程是不一樣的（被一道題坑過）。它的遞歸，是將左半部分先排好，再排右半部分，和教科書上的圖示是不一樣的過程，而非遞歸就和教科書上的一樣了。

//////////////////////////////////////歸併排序

void merge(int l1,int r1,int l2,int r2){
    int temp[n];
    int index=0;
    while(l1<=r1&&l2<=r2){
        if(a[l1]<a[l2])
            temp[index++]=a[l1++];
        else
            temp[index++]=a[l2++];
    }
    while(l1<=r1)temp[index++]=a[l1++];
    while(l2<=r2)temp[index++]=a[l2++];

    for(int i=0;i<index;i++)
        a[l1+i]=temp[i];
}

//遞歸
void mergesort(int left,int right){
    if(left>right)
        return;

    int mid=(left+right)/2;

    mergesort(left,mid);
    mergesort(mid+1;right);
    merge(left,mid,mid+1,right);
}

//非遞歸
void mergesort(){
    for(int step=2;step/2<n;step*=2){
        for(int i=0;i<n;i+=step){
            int end=min(i+step,n-1);
            int mid=i+step/2-1;
            merge(i,mid,mid+1,end);
        }
    }
}

//////////////////////////////////////歸併排序

三、快速排序

//////////////////////////////////////快速排序

int partition(int left,int right){
    int cur=a[left];
    while(left<right){
        while(left<right&&a[right]>temp)right--;
        a[left]=a[right];
        while(left<right&&a[left]<temp)left++;
        a[right]=a[left];
    }
    a[left]=cur;
    return left;
}

void quicksort(int left,int right){
    if(left<right){
        int mid=partition(left,right);
        quicksort(left,mid-1);
        quicksort(mid+1,right);
    }
}

//////////////////////////////////////快速排序

四、二分查找

二分查找本身非常簡單，但是它的逼近思想可以擴展出很多應用，例如求根號2的近似數等。

int binarysearch(int left,int right,int data){
    if(left>right)
        return -1;

    int mid=(left+right)/2;

    if(a[mid]==data)
        return mid;
    else if(data<a[mid])
        return binarysearch(left,mid-1);
    else
        return binarysearch(mid+1,right);
}

 

篇幅不夠，題目來湊！

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

#include<stdio.h>
#include<vector>
#include<algorithm>
#include<map>
#include<string>
#include<iostream>
#include<queue>
using namespace std;

struct node{
    string name,id;
    int grade;
};

int num;

vector<node> result;

bool cmp(node &a,node &b){
    return a.grade>b.grade;
}

int main(){
    cin>>num;

    for(int i=0;i<num;i++){
        string name,id;
        int grade;
        cin>>name>>id>>grade;
        node* temp=new node;
        temp->name=name;
        temp->id=id;
        temp->grade=grade;
        result.push_back(*temp);
    }

    int g1,g2;
    cin>>g1>>g2;

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

    bool flag=false;
    for(int i=0;i<result.size();i++){
        if(result[i].grade>=g1&&result[i].grade<=g2){
                flag=true;
                cout<<result[i].name<<" "<<result[i].id<<endl;
        }
    }

    if(!flag){
        cout<<"NONE"<<endl;
    }

    return 0;
}