23.Exploration-取绝对值排序

题目：

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 of N (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 in T (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.

输入：

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

输出：

<p>* Line 1: The maximum number of landmarks Bessie can visit.</p>

样例：

25 5
10
-3
8
-7
1

4

题目大意：求一下离着原点最近的点然后求规定时间内的参观的点的个数，从第一个点到第二个点的路程其实是第一个点坐标-第二个点坐标的绝对值

代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

#define MAX_N 50005

int A[MAX_N];
bool mycmp(int a, int b)
{
    return abs(a) < abs(b);
}

int main()
{
    int T, N;
    cin >> T >> N;
    for(int i = 0; i < N; i++)
        cin >> A[i];
    sort(A, A + N, mycmp);
    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];   ///上一个的位置
    }
    cout << ans << endl;
    return 0;
}