POJ 3258 River Hopscotch
来源:互联网 发布:淘宝设置发货地 编辑:程序博客网 时间:2024/05/23 01:59
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
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
Sample Input
25 5 2214112117
Sample Output
4
Hint
题意:在n个石头里面选择出m个石头,使得所有石头间距(包括河岸和石头)的最小值最大。类似于青蛙过河,二分。
#include <iostream>#include <stdio.h>#include <string>#include <cstring>#include <cmath>#include <algorithm>#define N 55555#define ll __int64using namespace std;ll a[N];ll L,n,m;int fun(int x){ int num=0; int f=0; for(int i=1;i<=n+1;i++) { if(x>=a[i]-a[f]) num++; else f=i; } if(num>m) return 1; else return 0;}int main(){ while(~scanf("%d%d%d",&L,&n,&m)) { for(int i=1;i<=n;i++) scanf("%d",&a[i]); a[0]=0; a[n+1]=L; sort(a+1,a+n+1); ll le=0,ri=L; ll mid; while(le<=ri) { mid=(le+ri)/2; if(fun(mid)) ri=mid-1; else le=mid+1; } cout<<le<<endl; } return 0;}
- poj 3258 River Hopscotch
- poj-3258 River Hopscotch
- poj-3258 River Hopscotch
- poj 3258 River Hopscotch
- poj 3258 River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258 River Hopscotch
- poj 3258 River Hopscotch
- POJ-3258-River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258 River Hopscotch
- poj 3258-River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258 River Hopscotch
- POJ 3258--River Hopscotch
- poj 3258 River Hopscotch
- Linux学习笔记(3-13)共享文件夹
- Use UIGestureRecognizer to handle single tap and double tap
- Spring配置文件详解 - applicationContext.xml文件路径
- mininet的中文教程学习
- c++控制台(非窗口,非MFC框架)里线程如何通过自定义消息通信
- POJ 3258 River Hopscotch
- DLL 中 DEF 文件的使用
- [LeetCode] 035. Search Insert Position (Medium) (C++)
- 2015/3/13这周总结
- 十六进制---十进制转换
- abc
- mysql的数据类型int、bigint、smallint 和 tinyint取值范围
- 01-1. 最大子列和问题
- Linux 云服务如何部署jdk环境