zoukankan      html  css  js  c++  java
  • River Hopscotch

    题目链接:http://poj.org/problem?id=3258

    River Hopscotch
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 22092   Accepted: 9138

    Description

    Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L).

    To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river.

    Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to rocks (0 ≤ M ≤ N).

    FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of Mrocks.

    Input

    Line 1: Three space-separated integers: LN, and M 
    Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position.

    Output

    Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks

    Sample Input

    25 5 2
    2
    14
    11
    21
    17

    Sample Output

    4

    Hint

    Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).

    Source

     
    题目大意:输入L N M  L代表所有的石头距离在起点的最大距离 N代表有N个石头  M代表你可以消去M个石头  问你消去M个石头之后   你走的每一步里面最小值的最大值是多少
    思路: 赤裸裸的二分 但是自己没想到 真的菜   其实自己的思路是  先把所有的点里原点的距离从小到大排序一遍,然后取最小的。 但是这样不容易处理啊  因为你取了一个的话  后面一个的距离和前面一个的
    距离就变了  如果要维护这一点的话  就算你记录了他们的下标  也改变了后面一个点和前面一个点的距离  然后呢?  你又得把新得到的数组重新从小到大排序  真的不好控制
    但是二分来解是真的快!
    二分的是答案    思路真的简单  但是没有想到     仔细想想,答案肯定在0-L之间的   判断一个值能否成立也很容易  太菜了。。。
    看代码:
     
    当初的梦想实现了吗,事到如今只好放弃吗~
  • 相关阅读:
    loadRunner运行场景时,事务数为0或是只显示添加的事务的数
    loadrunner12自带的机票预订服务,解决httpd: Could not reliably determine the server's fully qualified domain name 问题
    Loadrunner12的下载和安装
    【AI测试】人工智能 (AI) 测试--第二篇
    【AI测试】人工智能 (AI) 测试--开篇
    CVAT 用户指南
    List of devices attached 没有手机设备号 解决办法
    Pip安装Django超时(time out)解决方法
    ipsec_笔记
    cisco 2600(转)
  • 原文地址:https://www.cnblogs.com/caijiaming/p/10356054.html
Copyright © 2011-2022 走看看