POJ 3258 River Hopscotch 二分答案

River Hopscotch Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 6193 Accepted: 2685

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 D_i from the start (0 < D_i <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).

Source

USACO 2006 December Silver

題解

同樣是一個二分答案的題目，題意是講，一個長度為L的河，有N個墊腳石，現在需要去掉m個墊腳石，使得兩塊石頭之間的距離的最小值最大。把河的兩岸也看做墊腳石，但是不能去掉他們。

和前幾個題目一樣，通過二分答案區間就行了。但是這個題目和前幾道不同之處在於這個是修改搜索上界時更新res（前幾道題都是修改下界時更新）。最後返回res即可。

寫出來的效果就是代碼示例中的那個樣子了。

代碼示例

/****
	*@author    Shen
	*@title     poj 1064
	*/

#include 
#include 
#include 
#include 
using namespace std;

int n, k;
int l, v[50005];
int maxa = 0;

bool test(int x)
{
    int sum = 0, st = 0;
    for (int i = 1; i <= n + 1; i++)
    {
        if (v[i] - v[st] <= x)
            sum++;
        else
            st = i;
    }
    //printf("\t%s with x = %d, result is that sum = %d.\n", __func__, x, sum);
    return sum <= k;
}

int Bsearch(int l, int r)
{
    int res = r;
    while (l <= r)
    {
        int mid = (r + l) / 2;
        //printf("l = %d, r = %d, mid = %d, res = %d.\n", l, r, mid, res);
        if (test(mid))
             l = mid + 1;
        else
            res = min(res, mid), r = mid - 1;
    }
    return res;
}

void solve()
{
    maxa = l;
    v[0] = 0;
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &l);
        v[i] = l;
    }
    v[n + 1] = maxa;
    sort(v, v + n + 2);
    int ans = Bsearch(0, maxa);
    printf("%d\n", ans);
}

int main()
{
    while (~scanf("%d%d%d",&l, &n, &k))
        solve();
    return 0;
}