River Hopscotch

博客围绕River Hopscotch问题展开,讲述在一条有起始和终点石头的河流中,有若干中间石头,牛需从石头跳到石头到达终点。农夫John想移除部分石头以增加牛跳跃的最短距离,已知可移除石头数量,要求计算移除石头后牛跳跃的最大最短距离。

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River Hopscotch

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 20117 Accepted: 8363

Description

 

http://poj.org/problem?id=325

 

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 M rocks.

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).

 

题意

 

#include<stdio.h>
#include<string.h>
#include<cmath>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
const int mx=50010 ;
int a[mx];

int main()
{
	ios::sync_with_stdio(false);
	int l,n,m;
	cin>>l>>n>>m;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
	}
	n++;
	a[0]=0;
	a[n]=l;
	int q=0;///储存最短的答案
	sort(a+1,a+n+1);
	int li=0,re=l+l;
	while (li<re-1 ) { 
		int mid=li+re;///当距离为mid的时候,查看要去掉的石头有多少
		mid/=2;
		///cout<<mid<<endl;
		int flag=0;//计算当前剩余的第一颗石头是坐标
		int w=0;///表示搬走了w块石头
		for(int i=1;i<=n; )
		{
			if (a[i]-a[flag]<mid) {////如果第i块石头距离前一个合格石头的距离比较小,那么符合搬走的条件
				i++;///搬走
				w++;///块数加一
			}
			else
			{flag=i;
			i++;}
		}
		if(w>m)///如果w>m,说明不合格的石头太多,mid太大
		{
			re=mid;
		}
		else///否则,情况1,mid刚刚好,情况2,mid太小了,不符合情况的距离很少,所以要增加mid
		{
				li=mid;
		}
	}
	cout<<li<<endl;
	return 0;
}

 

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