1028 List Sorting (25 分)（排序）

Excel can sort records according to any column. Now you are supposed to imitate this function.

Input

Each input file contains one test case. For each case, the first line contains two integers N (<=100000) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then N lines follow, each contains a record of a student. A student’s record consists of his or her distinct ID (a 6-digit number), name (a string with no more than 8 characters without space), and grade (an integer between 0 and 100, inclusive).

Output

For each test case, output the sorting result in N lines. That is, if C = 1 then the records must be sorted in increasing order according to ID’s; if C = 2 then the records must be sorted in non-decreasing order according to names; and if C = 3 then the records must be sorted in non-decreasing order according to grades. If there are several students who have the same name or grade, they must be sorted according to their ID’s in increasing order.

Sample Input 1

3 1
000007 James 85
000010 Amy 90
000001 Zoe 60

Sample Output 1

000001 Zoe 60
000007 James 85
000010 Amy 90

Sample Input 2

4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98

Sample Output 2

000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60

Sample Input 3

4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 90

Sample Output 3

000001 Zoe 60
000007 James 85
000002 James 90
000010 Amy 90

生词

英文	解释
imitate	模仿
sort with	与...相符合
inclusive	包括的

题目大意：

根据c的值是1还是2还是3，对相应的列排序。第一列升序，第二列不降序，第三列不降序

分析：

注意，按照名称的不降序排序，因为strcmp比较的是ACSII码，所以A < Z。写cmp函数的时候return strcmp(a.name, b.name) <= 0; return语句返回的是true或者false的值，所以要写 <= 0 这样的形式。比较ACSII码的大小，strcmp(‘a’, ‘z’)返回负值，因为a<z a – z < 0
按照分数的不降序排序，a.score <= b.score

原文链接：https://blog.csdn.net/liuchuo/article/details/52144549

题解

以前（不知道啥时候做的）做的：

#include<bits/stdc++.h>

using namespace std;
struct student
{
    int id;
    char name[10];
    int grade;
} ;
bool cmp1(student a,student b)
{
    return a.id<b.id;
}
bool cmp2(student a,student b)
{
    if(strcmp(a.name,b.name)!=0)
        return strcmp(a.name,b.name)<=0;
    else
        return a.id<b.id;
}
bool cmp3(student a,student b)
{
    if(a.grade!=b.grade)
        return a.grade<=b.grade;
    else
        return a.id<b.id;
}
int main()
{
    int n,c;
    student stu[100000];
    scanf("%d %d",&n,&c);

    for(int i=0; i<n; i++)
    {
        scanf("%d %s %d",&stu[i].id,stu[i].name,&stu[i].grade);
    }
    if(c==1)
        sort(stu,stu+n,cmp1);
    else if(c==2)
        sort(stu,stu+n,cmp2);
    else
        sort(stu,stu+n,cmp3);
    //printf("Case %d:\n",c);
    for(int i=0; i<n; i++)
        printf("%06d %s %d\n",stu[i].id,stu[i].name,stu[i].grade);

    return 0;
}

现在做的：

#include <bits/stdc++.h>

using namespace std;
struct student
{
    long int id;
    string name;
    int score;
};
bool cmp1(student a,student b){
    return a.id<b.id;
}
bool cmp2(student a,student b){
    return a.name!=b.name?a.name<b.name:a.id<b.id;
}
bool cmp3(student a,student b){
    return a.score!=b.score?a.score<b.score:a.id<b.id;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,c;
    cin>>n>>c;
    vector<student> v(n);
    for(int i=0;i<n;i++){
        cin>>v[i].id>>v[i].name>>v[i].score;
    }
    if(c==1) sort(v.begin(),v.end(),cmp1);
    else if(c==2) sort(v.begin(),v.end(),cmp2);
    else sort(v.begin(),v.end(),cmp3);
    for(int i=0;i<n;i++)
    {
        printf("%06d %s %d\n",v[i].id,v[i].name.c_str(),v[i].score);
    }
}

柳神做的：

#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
const int maxn = 100001;
struct NODE {
    int no, score;
    char name[10];
}node[maxn];
int c;
int cmp1(NODE a, NODE b) {
    if(c == 1) {
        return a.no < b.no;
    } else if(c == 2) {
        if(strcmp(a.name, b.name) == 0) return a.no < b.no;
        return strcmp(a.name, b.name) <= 0;
    } else {
        if(a.score == b.score) return a.no < b.no;
        return a.score <= b.score;
    }
}
int main() {
    int n;
    scanf("%d%d", &n, &c);
    for(int i = 0; i < n; i++)
        scanf("%d %s %d", &node[i].no, node[i].name, &node[i].score);
    sort(node, node + n, cmp1);
    for(int i = 0; i < n; i++)
        printf("%06d %s %d\n", node[i].no, node[i].name, node[i].score);
    return 0;
}