[二分+贪心] 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
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).

为了判断所有可能出现的答案，使用二分法查找答案是否满足，问题就转换成了：给定步数mid，判断从N个点删除M个点以后是否能使每一个点的距离都大于mid。直接使用贪心，从头开始，每次贪心地选取距离大于mid的点，如果能选满N-M+2个，就表示可以完成，否则不能完成
二分法不断地搜索答案，就可以解决了。

#include <iostream>
#include <algorithm>
#include<cstdio>
using namespace std;
const int MAXN = 50050;
int L, N, M;
int D[MAXN];

bool C(int d)
{
    int last = 0;
    for (int i = 1; i < N - M; ++i)
    {
        int current = last + 1;
        while (current < N && D[current] - D[last] < d)
            ++current;
        if (current == N)
            return false;
        last = current;
    }
    return true;
}

int main()
{
    cin >> L >> N >> M;
    for (int i = 1; i <= N; ++i)
        cin >> D[i];
    D[N+1] = L;
    N = N+2;
    sort(D, D + N);
    int lb = 0, ub = L + 1;
    while (ub - lb > 1)
    {
        int mid = (lb + ub) / 2;
     //   printf("mid : %d\n",mid);
        if (C(mid))lb = mid;
        else ub = mid;
    }
    cout << lb << endl;
    return 0;
}