HUD2217

1.题目描述：

Visit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 855    Accepted Submission(s): 286

Problem Description

Wangye is interested in traveling. One day, he want to make a visit to some
different places in a line. There are N(1 <= N <= 2000) places located at points x1, x2, ..., xN (-100,000 ≤ xi ≤ 100,000). Wangye starts at HDU (x = 0), and he travels 1 distance unit in 1 minute . He want to know how many places he could visit at most, if he has T (1 <= T <= 200000 ) minutes.

 

Input

The input contains several test cases .Each test case starts with two number N and T which indicate the number of places and the time respectively. Then N lines follows, each line has a number xi indicate the position of i-th place.

 

Output

For each test case, you should print the number of places,Wangye could visit at most, in one line.

 

Sample Input

5 16-3-71108

 

Sample Output

4
Hint
In the sample, Wangye could visit (-3) first, then goes back to (0),  which costs him 6 minutes. Then he visits (1), (8), (10).  So he could visit 4 places at most in 16 minutes. :
 

 

Author

zjt

 

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2.题意概述：

给你N个坐标，初始的时候主人公位于点0处，所有点位于坐标轴上，问在时间T限制内，最多能够访问多少个点。一秒钟移动一个单位距离。

3.解题思路：

首先将N个点的坐标进行排序，然后设定l【】，表示坐标为负数的集合，设定r【】，表示坐标为正的集合，对应l【i】<l【i+1】（按照排序处理），同理r【i】<r【i+1】。
然后，暴力处理，不难分析出其有两种走法，暴力枚举一下，复杂度N方：
①选一个点，作为走到的左边的最远点，然后折返 ，向右走到时间耗用结束为止，维护过程中，最大值。
②选一个点，作为走到的右边的最远点，然后折返，向左走到时间耗用结束为止，维护过程中，最大值。

4.AC代码：

#include <stdio.h>#include <string.h>#include <cmath>#include <algorithm>#define maxn 22222using namespace std;int p[maxn], l[maxn], r[maxn];int main(){int n, t;while (scanf("%d%d", &n, &t) != EOF){memset(l, 0, sizeof(l));memset(r, 0, sizeof(r));for (int i = 0; i < n; i++)scanf("%d", &p[i]);sort(p, p + n);int cnt1 = 0, cnt2 = 0;for (int i = 0; i < n; i++){if (p[i] < 0)l[cnt1++] = p[i];elser[cnt2++] = p[i];}int ans = 0;for (int i = 0; i < cnt1; i++){int cnt = cnt1 - i;int time = t + 2 * l[i];if (time < 0)continue;for (int j = 0; j < cnt2; j++){int use = 0;if (j == 0)use = r[j];else use = r[j] - r[j - 1];if (time > 0 && use <= time){time -= use;cnt++;}else break;}ans = max(ans, cnt);}for (int i = 0; i < cnt2; i++){int cnt = i + 1;int time = t - r[i] * 2;if (time < 0)continue;for (int j = cnt1 - 1; j >= 0; j--){int use = l[j + 1] - l[j];if (time > 0 && time >= use){time -= use;cnt++;}else break;}ans = max(ans, cnt);}printf("%d\n", ans);}return 0;}