poj3258River Hopscotch（二分）



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 distanceD_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 toM 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 ofM 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 removingM rocks

Sample Input

25 5 22
#include<stdio.h>#include<iostream>#include<algorithm>using namespace std;#define N 50500int arr[N];int len;int n,m;int cheak(int d){    int x,y,num=0;    x=0;y=1;    for(;y<=n+1;)    {        if(arr[y]-arr[x]<d)        {            num++;            y++;        }        else        {            x=y;            y++;        }    }    if(num<=m) return 1;//这个判断想了好久都没想出简单的算法，这个简单算法是借鉴的大神的。    else return 0;}int main(){    while(cin>>len>>n>>m)    {        int i;        for(i=1;i<=n;i++)        {            scanf("%d",&arr[i]);        }        arr[n+1]=len;        arr[0]=0;        sort(arr,arr+n+1);        int be,en,mid,ans;        be=0;en=len;        while(be<=en)        {            mid=(be+en)/2;            if(cheak(mid)==1)            {                ans=mid;                be=mid+1;            }            else                en=mid-1;        }        printf("%d\n",ans);    }    return 0;}

14112117

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).

题意大致是：给定n个点，删除其中的m个点，其中两点之间距离最小的最大值。