URAL 1100. Final Standings (sort)

1100. Final Standings
Time limit:1.0 Second
Memory limit:16 MB
Old contest software uses bubble sort for generating final standings. But now, there is too many teams and that software works too slow. You is asked to write a program, which generates exactly the same final standings as old software, but fast. Inputthe first line of input contains only integer 1 < N≤150000-number of teams. Each of the next NLines contains, integers 1≤ ID≤107and 0≤ M≤100. ID-unique Number of Team, M-number of solved problems. Outputoutput should contain Nlines with integers IDand MOn each. Lines should be sorted by MIn descending order as produced by bubble sort (see below). Sample

input	Output
81 216 311 220 33 526 47 122 4	3 526 422 416 320 31 211 27 1

Notesbubble Sort Works following:
while (exists A[i] and A[i+1] such as A[i] < A[i+1]) do
   Swap(A[i], A[i+1]);

Test instructions: Sort by the second element, if the two are the same, do not change the original relative back and forth relationship.

Parse: Sort + record the initial position.

AC Code:

#include <cstdio> #include <algorithm>using namespace std;struct node{    int x, y, id;             ID record initial position}a[150005];int CMP (node AA, node bb) {    if (aa.y = = bb.y) return aa.id < bb.id;    return aa.y > Bb.y;} int main () {    #ifdef sxk        freopen ("In.txt", "R", stdin);    #endif//sxk    int n;    while (scanf ("%d", &n) ==1) {for        (int i=0; i<n; i++) {            scanf ("%d%d", &a[i].x, &a[i].y);            A[i].id = i;        }        Sort (A, a+n, CMP);        for (int i=0; i<n; i++) printf ("%d%d\n", a[i].x, A[I].Y);    }    return 0;}

URAL 1100. Final Standings (sort)