POJ 3618 - Exploration 不一样的排序解决办法

Description

Bessie is traveling on a road teeming with interesting landmarks. The road is labeled just like a number line, and Bessie starts at the "origin" (x = 0). A total ofN (1 ≤ N ≤ 50,000) landmarks are located at points x₁,x₂, ..., x_N (-100,000 ≤ x_i ≤ 100,000). Bessie wants to visit as many landmarks as possible before sundown, which occurs inT (1 ≤ T ≤ 1,000,000,000) minutes. She travels 1 distance unit in 1 minute.

Bessie will visit the landmarks in a particular order. Since the landmarks closer to the origin are more important to Farmer John, she always heads for the unvisited landmark closest to the origin. No two landmarks will be the same distance away from the origin.

Help Bessie determine the maximum number of landmarks she can visit before the day ends.

Input

* Line 1: Two space-separated integers: T and N
* Lines 2..N+1: Line i+1 contains a single integer that is the location of theith landmark: x_i

Output

* Line 1: The maximum number of landmarks Bessie can visit.

Sample Input

25 510-38-71

Sample Output

4

应用abs（）排序直接求结果

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <queue>#include <set>#include <map>#include <algorithm>using namespace std;typedef pair<int, int> P;//#define LOCAL#define INF 0x3f3f3f3f#define MAX_N 50005int A[MAX_N];bool mycmp(int a, int b){    return abs(a) < abs(b);}int main(){#ifdef LOCALfreopen("b:\\data.in.txt", "r", stdin);#endif    int T, N;    cin >> T >> N;    for(int i = 0; i < N; i++)        cin >> A[i];    sort(A, A + N, mycmp);//    for(int i = 0; i < N; i++)//        cout << A[i] << " ";    long long dis = 0;    int ans = 0;    int las = 0;    for(int i = 0; i <= N; i++)    {        dis += abs(A[i] - las);        ans++;        if(dis > T)        {            ans--;            break;        }        las = A[i];    }    cout << ans << endl;    return 0;}