River Hopscotch(二分最大值)

WPF实战:利用Live Charts实现高性能动态曲线可视化 本文详细介绍了如何在WPF中使用Live Charts实现高性能动态曲线可视化,特别适用于工业监控和金融分析等实时数据场景。通过增量渲染、内存复用和硬件加速等核心技术,Live Charts能够高效处理高频数据更新,显著提升图表流畅度。文章提供了从基础实现到工业级优化的完整解决方案,帮助开发者应对各种性能挑战。 阅读详情

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 M 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 M rocks.

Input
Line 1: Three space-separated integers: L, N, 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).

给出一堆石头坐标,一只青蛙xjb来回跳,最多略过m个石头,问他跳的过程中,最短的一次跳跃距离最大能是多少。

直接二分枚举,判断也很简单,每次跳的时候略过石头数加起来,与m比较。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<utility>
#include<sstream>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
int l,m,n;
int a[50010];
bool check(int x)
{
    int pre = 0;
    for(int i = 1;i <= n - m;i++)
    {
        pre = lower_bound(a + pre,a + 1 + n,a[pre] + x) - a;
        if(pre > n)return false;
    }
    if(l - a[pre] < x)return false;
    return true;
}
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\巍巍\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\巍巍\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    scanf("%d%d%d",&l,&n,&m);
    a[0] = 0;a[n + 1] = l;
    for(int i = 1;i <= n;i++)
        scanf("%d",&a[i]);
    sort(a + 1,a + 1 + n);
    int l = 0,r = inf;
    while(r - l > 1)
    {
        int mid = (l + r)/2;
        if(check(mid))
            l = mid;
        else
            r = mid;
    }
    printf("%d\n",l);
    return 0;
}
鸿蒙平板PC应用开发:在termux中安装ubuntu 打开AVNC点击右上角的设置图标,然后点击 输入 , 点击手势栏目中的 单指轻扫 ,选择滚动远程内容 阅读详情

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River Hopscotch [二分]

题意:马跳石头,给定石头的位置,求拿掉m个石头后,马到目的地过程中跳跃的最小距离的最大值。 思路:给定的石头位置,初始0,目的地L, 排个序。然后二分最小距离找最大值。 二分的单调性:如果马的最小一步能跨越ll,那么对于任意的li>ll_i>l作为最小步长都是合理的。 二分的check(mid):如果当前最小步长mid满足仅有cnt<=mcnt <=m个小于mid的长度时,以当前mid作为最小

Qer的博客 453

River Hopscotch

题目链接:https://cn.vjudge.net/contest/279225#problem/C 题目大意:就是一群自以为很聪明的牛要过河,它们的主人也是可怕,还要给它们举办节日,嗐……但是呢,今年主人想要给他们增加点难度,所以要去掉几块石头,让所有石头之间的最小距离最大,然后呢起始点的石头和终点的石头不能移走,可能是怕他们掉下去就不能产奶或者吃牛肉了吧,话说牛肉好好吃欸哈哈。就是一个最小值...

L_X_ 301

POJ3258 River Hopscotch(二分)

题目链接:https://vjudge.net/problem/POJ-3258 题意: 有一条河长为L,河中间有n块石头,算上开始和结尾一共n+2个,现在问去掉m块石头,问最短距离中的最大值是多少? 分析: 二分寻找一个值,使得可以移除m个石头。 code: // #include <bits/stdc++.h> #include <cstdio> #inc...

成龙大侠的博客 338

River Hopscotch (二分法)

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 ro...

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