1028 List Sorting (25 分)

1028 List Sorting (25 分)

 

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

Input Specification:

Each input file contains one test case. For each case, the first line contains two integers N (≤) 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 Specification:

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

三种排序方式，简单排序算是

#include <bits/stdc++.h>
using namespace std;
struct Node
{
    string id;
    string name;
    int sorce;
}node[100005];
bool cmp(Node a, Node b){
    return a.id < b.id;
}

bool cmp1(Node a, Node b){
    for(int i = 0; i < max(a.name.length(), b.name.length()); i++){
        if(min(a.name.length(), b.name.length()) == i)
            return a.name < b.name;
        if(a.name[i] == b.name[i])
            continue;
        return a.name[i] < b.name[i];
    }
    return a.id < b.id;
}

bool cmp2(Node a, Node b){
    if(a.sorce == b.sorce)
        return a.id < b.id;
    return a.sorce < b.sorce;
}

int n,m;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    for(int i = 0 ; i < n; i++){
        cin >> node[i].id>>node[i].name>>node[i].sorce;
    }
    if(m == 1){
        sort(node,node+n,cmp);
    }else if(m == 2){
        sort(node,node+n,cmp1);
    }else{
        sort(node,node+n,cmp2);
    }
    for(int i = 0; i < n; i++){
        cout <<node[i].id<<" "<<node[i].name<<" "<<node[i].sorce<<endl;
    }
    return 0;
}