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

8
1 2
16 3
11 2
20 3
3 5
26 4
7 1
22 4

3 5
26 4
22 4
16 3
20 3
1 2
11 2
7 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;
}