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  • 1028 List Sorting (25)

    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


     1 #include<iostream>
     2 #include<algorithm>
     3 #include<vector>
     4 #include<cstring>
     5 using namespace std;
     6 struct node{
     7   char name[9];
     8   int id, score;
     9 };
    10 
    11 int c;
    12 bool cmp(node a, node b){
    13   if(c==1) return a.id<b.id;
    14   if(c==2) return strcmp(a.name, b.name)==0 ? a.id<b.id : strcmp(a.name, b.name)<0;
    15   if(c==3) return a.score==b.score? a.id<b.id : a.score<b.score;
    16 }
    17 int main(){
    18   int n, i;
    19   cin>>n>>c;
    20   vector<node> v(n);
    21   for(i=0; i<n; i++) scanf("%d %s %d", &v[i].id, v[i].name, &v[i].score);
    22   sort(v.begin(), v.end(), cmp);
    23   for(i=0; i<n; i++) printf("%06d %s %d
    ", v[i].id, v[i].name, v[i].score);
    24   return 0;
    25 }
     
    有疑惑或者更好的解决方法的朋友,可以联系我,大家一起探讨。qq:1546431565
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  • 原文地址:https://www.cnblogs.com/mr-stn/p/9163559.html
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