A sort (nyoj8) (Simple sort)

A sort time limit: theMs | Memory Limit:65535KB Difficulty:3

Describe

Now there are a lot of rectangles, each rectangle has a number, this number can be repeated, also know that the width and length of the rectangle, the number, length, width are all integers; now it's required to sort by a bit (the default collation is small to large);

1. Sort by number from small to large

2. For the numbers equal to the rectangle, according to the oblong long sort;

3. If the numbers and lengths are the same, sort by the width of the rectangle;

4. If the number, length, and width are the same, only one rectangle is reserved for sorting, the extra rectangles are removed, and the final sequence displays all rectangles in the specified format;

Input

the first line has an integer 0<n<10000, which indicates that there is an n set of test data next.
The first line of each group has an integer 0<m<1000, which indicates that there is a m rectangle;
The next M-line, each line has three numbers, the first number represents the number of rectangles,

The second and third numeric representations are long, the values are small, the representation is wide, equal
Description This is a square (data contract length width and number are less than 10000);

Output

sequential output The number length of all eligible rectangles for each set of data

Sample input

181 1 11 1 11 1 21 2 11 2 22 1 12 1 22 2 1

Sample output

1 1 11 2 11 2 22 1 12 2 1

Source

Classic Topics

#include <stdio.h> #include <algorithm>using namespace std;struct st{int xh;int c,k;} Data[1010];int CMP (ST A,st B)//sequencing key! {if (A.XH!=B.XH) return A.xh<b.xh;else if (A.C!=B.C) return A.c<b.c;else if (A.K!=B.K) return A.K<B.K;} int main () {int test,i,n,t;scanf ("%d", &test), while (test--) {scanf ("%d", &n), and for (i=0;i<n;i++) {scanf ("%d%d %d ", &data[i].xh,&data[i].c,&data[i].k);//Pay attention to the length-width size. if (DATA[I].C<DATA[I].K) {t=data[i].c;data[i].c=data[i].k;data[i].k=t;}} Sort (data,data+n,cmp); for (i=0;i<n;i++) {if (!) ( DATA[I].XH==DATA[I+1].XH&&DATA[I].C==DATA[I+1].C&&DATA[I].K==DATA[I+1].K)) printf ("%d%d%d\n", DATA[I].XH,DATA[I].C,DATA[I].K);}} return 0;}

A sort (nyoj8) (Simple sort)