hrbust 1006 River Hopscotch【二分查找】好题-优快云博客

本文介绍了一个有趣的问题：如何通过移除部分河中的石头来最大化最小跳跃距离，旨在帮助农夫约翰解决其牛群过河的挑战。文章详细阐述了解决方案的思路与实现过程。

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River Hopscotch

Time Limit: 2000 MS

Memory Limit: 65536 K

Total Submit: 253(111 users)

Total Accepted: 113(91 users)

Rating:

Special Judge: No

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

For each test case :
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.
Process to the end of file.

Output

For each test case :
Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks

Sample Input

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：从起点到终点距离为L长度的河，其中有N块石头，必须拆除m块石头，在这样的情况下，求从起点蹦到终点需要蹦的最小距离。

思路：

1、枚举这个最小距离，因为最小距离的枚举范围太大，所以我们使用二分匹配算法来优化时间复杂度。

2、根据枚举出来的最小距离，对于排序之后的每一块石头，如果从上一块石头蹦到这一块石头的距离小于我们枚举出来的mid，那么这一块石头就要被拆除，然后看必须拆除的石头数是否大于m，如果大于m，那么当前情况不合法，显然我们枚举出来的值大了，然后继续二分。

3、我们记录的是可行解，并不能直接输出mid。

Ac代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int a[5000000];
int main()
{
    int lengh,n,m;
    while(~scanf("%d%d%d",&lengh,&n,&m))
    {
        a[0]=0;
        a[n+1]=lengh;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a,a+n+1);
        int l=0;
        int r=lengh;
        int mid;
        int ans=0;
        while(r-l>=0)
        {
            mid=(l+r)/2;
            int cont=0;
            int pre=0;
            for(int i=1;i<=n+1;i++)
            {
                if(a[i]-a[pre]<mid)cont++;
                else pre=i;
            }
            if(cont>m)
            {
                r=mid-1;
            }
            else
            {
                l=mid+1;
                ans=mid;
            }
        }
        printf("%d\n",ans);
    }
}