URAL 1100. Final Standings （排序）

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 are too many teams and that software works too slow. You are asked to write a program, which generates exactly the same final standings as old software, but fast.

Input

The first line of input contains only integer 1 < N ≤ 150000 — number of teams. Each of the next Nlines contains two integers 1 ≤ ID ≤ 107 and 0 ≤ M ≤ 100. ID — unique number of team, M — number of solved problems.

Output

Output should contain N lines with two integers ID and M on each. Lines should be sorted by M in descending order as produced by bubble sort (see below).

Sample

inputoutput

81 216 311 220 33 526 47 122 4

3 526 422 416 320 31 211 27 1

Notes

Bubble sort works following way: 
while (exists A[i] and A[i+1] such as A[i] < A[i+1]) do
   Swap(A[i], A[i+1]);

题意：按第二元素排序，若两者相同，不改变两者原来的相对前后关系。

解析：sort + 记录一下初始位置即可。

AC代码：

#include <cstdio>#include <algorithm>using namespace std;struct node{    int x, y, id;             //id记录初始位置}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;}