hdu 2217 Visit

最新推荐文章于 2019-01-06 21:10:47 发布

原创最新推荐文章于 2019-01-06 21:10:47 发布 · 468 阅读

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本文介绍了一道关于旅行者在限定时间内能访问多少地点的问题，并给出了解决方案。通过将地点进行排序并采用不同的旅行策略，如一直向左、一直向右、先左后右或先右后左等，来确定在规定时间内可以访问的最大地点数量。

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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
-7
1
10
8
Sample Output
4

题目大意：在数轴上给n个点，让你从点0开始走，算出在规定时间T内能拜访最多点的个数。

思路：开始我居然用dfs去写，虽然意料到会超时，但是我还是死脑筋的去想减支，后来发现完全可以不需要dfs！！

从0点开始走，要么左要么右，而且走到最右或者最左那么0到最左最右的点肯定拜访过了，当你望左走了后，如果你在往右是可以的，但是不能再往左，既不能左右左，因为这是徒劳的走法，路程肯定多，（仔细想一想）。所以走法只有：4种：一直向左，一直向右，先左后右，先右后左。

<span style="font-family:Arial;">#include <iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int n,T,a[2005];
int main()
{
    int s,q,p,ma,x,y;
    while(scanf("%d %d",&n,&T)!=EOF)
    {
        a[0]=0;
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        sort(a,a+n+1);//从小到大排序，
        s=lower_bound(a,a+n+1,0)-a;//找到点0
        ma=0;
        for(q=s;q>=0;q--)//q是往左
        {
            for(p=s;p<=n;p++)//p是往右
            {
                x=-a[q],y=a[p];
                if(2*x+y<=T&&p-q>ma) ma=p-q;//左右
                if(2*y+x<=T&&p-q>ma) ma=p-q;//右左
            }
        }
        cout<<ma<<endl;
    }
    return 0;
}</span>