Problem 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 of N (1 ≤ N ≤ 50,000) landmarks are located at points x1, x2, ..., xN (-100,000 ≤ xi ≤ 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.

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 the ith landmark: xi

Output

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

Sample Input

25 5

10
-3
8
-7
1

Sample Output

4

題意：一條數(shù)軸上，給n個點，從原點出發(fā)，優(yōu)先達到距離原點最近的點，求t秒內到達的點的最大數(shù)量。

思路：點的坐標有正有負，升序排序時按絕對值大小排序，然后計算t秒內到達的最大數(shù)量即可。

Source Program

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 100001 #define MOD 2520 #define E 1e-12 using namespace std; int a[N]; bool cmp(int x,int y) {return abs(x)<abs(y); } int main() {int t,n;scanf("%d%d",&t,&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);sort(a+1,a+1+n,cmp);int cnt=0;int use=0;//已用時間int k=1;if(a[1]==0){cnt=1;k++;}for(int i=k;i<=n;i++){int time=abs(a[i]-a[i-1]);if(time<=t-use)//若時間足夠，則前往下一點{cnt++;use+=time;//累計已用時間}elsebreak;}printf("%d\n",cnt);return 0; }

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