River Hopscotch POJ

River Hopscotch

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 14623 Accepted: 6210

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 2214112117

Sample Output

4

题目大意：一群牛要过河，河中有许多石墩，FJ为了锻炼牛的跳跃力，将其中的m个石墩摘掉，求所有情况中最大的跳跃最小值。（最大化最小值）

分析：跳跃值通过二分来枚举。对于某一个mid，如果可以摘除的石墩超过m的话，说明牛的跳跃值太大了，并不是最小跳跃值。如果可以摘除的石墩刚好等于m的话，也不一定是最小的。如果小于m的话，说明mid偏小。

代码如下：

/* Author:kzl */#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;const int maxx = 50000 + 500;int ll,n,m;int cun[maxx];bool flag(int num){//判断当最小距离为num的时候是否要拿走的石头多于M    int cnt = 0;    int last = 0;cun[0] = 0;cun[n+1] = ll;    for(int i=1;i<=n+1;i++){        if(cun[i]-cun[last]<num){            cnt++;        }        else {            last = i;        }    }if(cnt>m)return false;else return true;}int main(){while(scanf("%d%d%d",&ll,&n,&m)!=EOF){    for(int i=1;i<=n;i++)scanf("%d",&cun[i]);    sort(cun+1,cun+n+1);    int l = 1,r = ll,mid = 0;    while(l<=r){        mid = (r+l)/2;        if(flag(mid))l = mid + 1;//满足条件，说明mid可以作为最小值。        else r = mid - 1;    }    printf("%d\n",l-1);}return 0;}