River Hopscotch
poj 3258
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, Lunits 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 2
2
14
11
21
17
Sample Output
4
Hint
题意：
一条长l的河中，有n个垫脚石，现在给出它们距离起始点的距离，要求移除其中的m块，使得间距最小的两块石头之间的距离最大。
题解：
专题练习是二分，当然这是一个二分题，但是二分什么呢~果然还是太菜了啊啊啊啊啊。。。今天学长们青岛站ICPC打了历史最佳成绩，羡慕~还是要好好努力啊！！
我们已知河流的长度，那么两个石头之间的最小距离的最大值一定不会大于河流长度，于是我们可以二分长度，找到一个合适的长度，使得去掉的石子数正好是m个，而且该长度是最小长度的最大值。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 using namespace std;
6 const int maxn=50010;
7 int a[maxn],m,n;
8 int len;
9 bool check(int x)
10 {
11 int sum=0;
12 int t=0;
13 for(int i=1;i<=n+1;i++)
14 {
15 if(a[i]-a[t]<x)
16 {
17 sum++;
18 if(sum>m)
19 return false;
20 }
21 else
22 {
23 t=i;
24 }
25 }
26 return true;
27 }
28 int main()
29 {
30 scanf("%d%d%d",&len,&n,&m);
31 a[0]=0;
32 for(int i=1;i<=n;i++)
33 scanf("%d",&a[i]);
34 sort(a+1,a+1+n);
35 a[n+1]=len;
36 int l=0,r=len;
37 int ans=-1;
38 while(r>=l)
39 {
40 int mid=(r+l)/2;
41 if(check(mid))
42 {
43 ans=mid;
44 l=mid+1;
45 }
46 else
47 {
48 r=mid-1;
49 }
50 }
51 printf("%d
",ans);
52 return 0;
53 }