River Hopscotch(二分POJ3258)

River Hopscotch
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 9263 Accepted: 3994

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).
题意:
一条河中有一些石头,给你这些石头距离起点的距离,和河的宽度,问移去m块石头后,剩余石头之间距离的最小距离的最大值.
思路:
二分石头的最小距离,判断这个距离需要移动的石头的数目,不断的进行二分找到一个最大值.

#include <set>
#include <map>
#include <list>
#include <stack>
#include <cmath>
#include <vector>
#include <queue>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define PI cos(-1.0)
#define RR freopen("input.txt","r",stdin)
using namespace std;
typedef long long LL;
const int MAX = 50010;
LL  Arr[MAX];
int n,m;
LL l;
bool Judge(LL s)
{
    LL ans=0;
    int num=0;
    for(int i=0;i<=n;i++)
    {
        if(ans+Arr[i+1]-Arr[i]>=s)
        {
            ans=0;
        }
        else
        {
            num++;
            ans+=(Arr[i+1]-Arr[i]);
        }
        if(num>m)
        {
            return false;
        }
    }
    return true;
}

int main()
{

    while(~scanf("%lld %d %d",&l,&n,&m))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&Arr[i]);
        }
        Arr[0]=0;
        Arr[n+1]=l;
        sort(Arr,Arr+n+1);
        LL L=0,R=l;
        LL ans=0;
        while(L<=R)
        {
            LL mid=(L+R)>>1;
            if(Judge(mid))
            {
                ans=max(ans,mid);
                L=mid+1;
            }
            else
            {
                R=mid-1;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}