二分--POJ-3258

POJ-3258，二分

题目

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).

思路

题意，移除m个石子后，使相邻石子间的最短间距最大化。

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <sstream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iomanip>
#include <stack>

using namespace std;

typedef long long LL;
const int INF = 0x3f3f3f3f;
const int N = 50005;
const int MOD = 1e9 + 9;

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

int main()
{
    int L, n, m;
    cin >> L >> n >> m;
    int a[N];
    a[n] = L;
    for(int i = 0;i < n;++i)
        cin >> a[i];
    sort(a, a + n + 1);
    int l = 0, r = L;
    //在L和原来石子间的最短距离间二分，这里偷懒，在L到0间二分
    while(l <= r)
    {
        int mid = (l + r) >> 1;
        int cnt = 0, s = 0;//计算移除的石子个数
        for(int i = 0;i <= n;++i)
        {
            if(mid >= a[i] - s)//满足这条不等式时，即表示石子i可移除，同时累加一段距离至不能移除
                cnt++;
            else//累加至一段距离到不满足以上不等式时，就要从新的起点开始累加
                s = a[i];
        }
        if(cnt <= m)//这里要取=，因为当我们取走最后一个石子的时候，因为石子被取走，所以不能得出答案
            //而这时候就是让cnt==m，把l加到最大值，就是刚好不能移除多一个石子的距离，就是答案了
            l = mid + 1;
        else
            r = mid - 1;
    }
    cout << l << endl;
    return 0;
}

参考

https://blog.csdn.net/lyy289065406/article/details/6648558