# poj 3258 River Hopscotch

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

题意：给你一条路，然后在路的起点和终点上都一个石子，然后给你n个石子，告诉你这些石子的位置，然后你最多可以删除k个石子，问这些石子的最大化最小距离是多少，起点和终点的石子是不可以删除的。

思路：我们可以二分答案，然后用去检查这个距离是是否满足最多删除k个石子，找到一个最右端的值就是答案。

#include<iostream> #include<cstdio> #include<algorithm> #include<map> #include<cstring> #include<map> #include<set> #include<queue> #include<stack> #include<cmath> using namespace std; const double eps = 1e-10; const int inf = 0x3f3f3f3f, N = 1e5 + 10, Mod = 100000007; typedef long long ll; int a[N], n, m, L; int cheak(int x) { int cnt = 0; for(int i = 0; i<=n; i++) { int j = i+1; while(j<=n+1 && a[j] - a[i] < x) { cnt++; j++; } i = j-1; } return cnt <= m; } int main() { while(cin>>L>>n>>m) { for(int i =1; i<=n; i++) scanf("%d", &a[i]); sort(a+1, a+n+1); a[0] = 0, a[n+1] = L; int l = 0, r = L, mid; while(l<=r) { mid = (l+r)>>1; if(cheak(mid)) l = mid + 1; else r = mid - 1; } cout<<r<<endl; } return 0; }

最后更新于 2017-03-26 21:58:49 并被添加「二分」标签，已有 360 位童鞋阅读过。

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