POJ-3258

River Hopscotch

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16074		Accepted: 6754

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).

题意：

过一个长度为L的河，其中有n块石头，从中剔除m块， 求最后石头之间的最小距离的最大值。

可以使用二分答案的方法，从L-0开始。

AC代码：

//#include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int n,l,m,mid,ans;
int a[50010];

bool search(int t){//计算当前答案是否符合m 
    int sum=0,temp=0;
    for(int i=1;i<=n+1;i++){
        if(a[i]-a[temp]<t){
            sum++;
            if(sum>m){//当小于mid的距离数大于m，则说明此答案不合适 
                return false;
            }
        }
        else{
            temp=i;
        }
    }
    return true;
}

int main() {
    ios::sync_with_stdio(false);
    while(cin>>l>>n>>m){//输入语句，此处等同于printf 
        for(int i=1;i<=n;i++){
            cin>>a[i];
        }
        a[0]=0;//起始点也看做一个石头 
        a[n+1]=l;//终点 
        sort(a,a+n+1);//从0~n+1排序 
        int maxn=l;
        int minn=0;//最长距离即为此时maxn-minn，答案只能在这个区间内 
        while(maxn-minn>=0){ 
            mid=(maxn-minn)/2+minn;//二分 
            if(search(mid)){
                ans=mid;//记录答案，最后记录的值即为答案 
                minn=mid+1;
            }
            else{
                maxn=mid-1;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}