# [POJ 3258] River Hopscotch【二分答案】

## Problem:

Time Limit: 2000MS |
Memory Limit: 65536K |

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 *D _{i}* from the start (0 <

*D*<

_{i}*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

*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

*M*rocks

Sample Input

25 5 2 2 14 11 21 17

Sample Output

4

Hint

Source

## Solution:

首先将坐标**排序**，并将终点等价于 Rock[N + 1]，便于处理。

然后**二分单次跳跃最小距离**，check 时模拟跳跃，若当前距离小于最小距离，则**必须移除**该石块。

## Code: O(nlogn) [360K, 125MS]

#include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; int L, N, M, Rock[50005]; inline bool check(const int &minDis){ int curPos = 0, removeCnt = 0; for(register int i = 1;; i++){ while(i <= N + 1 && Rock[i] - curPos < minDis) removeCnt++, i++; if(i > N + 1) break; if(removeCnt > M) return 0; // Improve efficiency curPos = Rock[i]; } return removeCnt <= M; } int main(){ scanf("%d%d%d", &L, &N, &M); for(register int i = 1; i <= N; i++) scanf("%d", Rock + i); sort(Rock + 1, Rock + N + 1), Rock[N + 1] = L; // Simplify handling int lft = 1, rt = L; while(lft < rt){ int mid = lft + rt + 1 >> 1; if(check(mid)) lft = mid; else rt = mid - 1; } printf("%d\n", lft); return 0; }

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