http://www.lydsy.com/JudgeOnline/problem.php?id=1650

看到数据和最小最大时一眼就是二分。。。

但是仔细想想好像判断时不能贪心?

然后看题解还真是贪心。。囧。

原来是之前我脑残了。

。。。

贪心很简单

排序后。

当前点到之前的点的距离<m就累计(相当于删掉这个点,为什么呢?因为这个点假设last到的不是0,那么这个点删了后,因为后边的点的距离大于它,假设后边的点距离减去这个点的距离也是<m,那么显然删去这个点可以得到2个不用删去的点(否则一定要删去这两个点,这样就不是最优了)

否则之前的点变成当前点。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; } const int N=50005;
int a[N], n, m, L;
bool check(int s) {
int last=0, tot=0;
for1(i, 1, n) {
if(a[i]-a[last]<s) {
++tot;
if(tot>m) return false;
}
else last=i;
}
return true;
}
int main() {
read(L); read(n); read(m);
for1(i, 1, n) read(a[i]);
sort(a+1, a+1+n);
a[n+1]=L;
int l=0, r=L;
while(l<=r) {
int m=(l+r)>>1;
if(check(m)) l=m+1;
else r=m-1;
}
print(l-1);
return 0;
}

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.

数轴上有n个石子,第i个石头的坐标为Di,现在要从0跳到L,每次条都从一个石子跳到相邻的下一个石子。现在FJ允许你移走M个石子,问移走这M个石子后,相邻两个石子距离的最小值的最大值是多少。

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

5 rocks at distances 2, 11, 14, 17, and 21. Starting rock at position
0, finishing rock at position 25.

Sample Output

4

HINT

移除之前,最短距离在位置2的石头和起点之间;移除位置2和位置14两个石头后,最短距离变成17和21或21和25之间的4.

Source

【BZOJ】1650: [Usaco2006 Dec]River Hopscotch 跳石子(二分+贪心)的更多相关文章

  1. bzoj 1650: [Usaco2006 Dec]River Hopscotch 跳石子【贪心+二分】

    脑子一抽写了个堆,发现不对才想起来最值用二分 然后判断的时候贪心的把不合mid的区间打通,看打通次数是否小于等于m即可 #include<iostream> #include<cst ...

  2. bzoj 1650: [Usaco2006 Dec]River Hopscotch 跳石子

    1650: [Usaco2006 Dec]River Hopscotch 跳石子 Time Limit: 5 Sec  Memory Limit: 64 MB Description Every ye ...

  3. BZOJ 1650 [Usaco2006 Dec]River Hopscotch 跳石子:二分

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1650 题意: 数轴上有n个石子,第i个石头的坐标为Di,现在要从0跳到L,每次条都从一个石 ...

  4. bzoj1650 [Usaco2006 Dec]River Hopscotch 跳石子

    Description Every year the cows hold an event featuring a peculiar version of hopscotch that involve ...

  5. BZOJ 1717: [Usaco2006 Dec]Milk Patterns 产奶的模式 [后缀数组]

    1717: [Usaco2006 Dec]Milk Patterns 产奶的模式 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 1017  Solved: ...

  6. Bzoj 1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐 深搜,bitset

    1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 554  Solved: 346[ ...

  7. BZOJ 1649: [Usaco2006 Dec]Cow Roller Coaster( dp )

    有点类似背包 , 就是那样子搞... --------------------------------------------------------------------------------- ...

  8. BZOJ 1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐( dfs )

    直接从每个奶牛所在的farm dfs , 然后算一下.. ----------------------------------------------------------------------- ...

  9. BZOJ 1717: [Usaco2006 Dec]Milk Patterns 产奶的模式( 二分答案 + 后缀数组 )

    二分答案m, 后缀数组求出height数组后分组来判断. ------------------------------------------------------------ #include&l ...

随机推荐

  1. 【Linux】用户权限设置,配合FTP访问

    转载自: http://blog.csdn.net/fengeh/article/details/16819563 领导需求,需要创建用户,并允许其增删改,却又要求其只能在自己的访问目录内,不能去别的 ...

  2. SQL 防止注入

    var strsql = "insert into Staff_Answer (ExamTitleID,QuestionsID,MultipleChoice,RightOption,Answ ...

  3. window.open()的所有参数列表

    http://www.cnblogs.com/meil/archive/2006/07/28/462459.html[1.最基本的弹出窗口代码] 其实代码非常简单: <SCRIPT LANGUA ...

  4. JAVA中的static方法

    static表示“全局”或者“静态”的意思,用来修饰成员变量和成员方法,也可以形成静态static代码块,但是Java语言中没有全局变量的概念. 被static修饰的成员变量和成员方法独立于该类的任何 ...

  5. 摘:C#压缩文件

    [[[[C#压缩文件]]]] 方法1: //[filepath想要压缩文件的地址] //[zippath输出压缩文件的地址] private void GetFileToZip(string file ...

  6. C#四舍五入保留一位小数

    DateTime d1 = hrStaff.DateJoin; DateTime d2 = DateTime.Now; TimeSpan d3 = d2.Subtract(d1); ; //int i ...

  7. blender, 创建多边形面片

    按a键清除所有选择,进入Edit Mode,选vertex select方式.然后按住control,使用MLB连续画多个顶点,形成一个多边形,如图所示: 然后同时选中两个端点,点Make Edge/ ...

  8. ui-router(三)controller与template

    这篇就是在以前的基础上,把客户端angular.js 负责的部分整体串起来演示一下. 我们按照angular执行顺序来做前提准备: (1)Client 根目录下 index.html 首先加载angu ...

  9. [svc]sudo su权限案例

    一 控制sudo 允许执行所有命令,排除某几个命令(带参数) lanny ALL=(ALL) NOPASSWD:ALL, !/bin/su - root, !/usr/sbin/visudo 如果需要 ...

  10. PaaS 平台的网络需求

    在使用 Docker 构建 PaaS 平台的过程中,我们首先遇到的问题是需要选择一个满足需求的网络模型: 让每个容器拥有自己的网络栈,特别是独立的 IP 地址 能够进行跨服务器的容器间通讯,同时不依赖 ...