（POJ - 3258）River Hopscotch

（POJ - 3258）River Hopscotch

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 15177 Accepted: 6440

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

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).

题目大意：一群牛要过河，河中有n块石头，FJ为了锻炼牛的跳跃能力，决定拿掉m块石头，使得相邻两块石头之间的最小距离尽可能大。（最大化最小值）

思路：这种类型的题目很容易想到二分。显然答案在[1,L]之间，我们通过二分来枚举，对于每一个值，去判断一下可以拿走的石头数量cnt。如果cnt<=m，则说明二分的这个mid偏小了，如果大于m则说明二分的这个mid偏大了。

#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn=50005;
int a[maxn];
int l,n,m;

bool check(int x)
{
    int cnt=0,tmp=0;
    for(int i=1;i<=n+1;i++)
    {
        if(a[i]-tmp<x) cnt++;
        else tmp=a[i];
    }
    if(cnt<=m) return true;//说明二分的答案偏小了 
    else return false;
}

int main()
{
    while(scanf("%d%d%d",&l,&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++) scanf("%d",a+i);
        a[0]=0;
        a[n+1]=l;
        sort(a,a+n+2);
        int lo=1,hi=l,mid;
        int ans;
        while(lo<=hi)
        {
            mid=(lo+hi)>>1;
            if(check(mid))
            {
                lo=mid+1;
                ans=mid;
            }
            else hi=mid-1; 
        }
        printf("%d\n",ans);
    }
    return 0;
}