zhber 有好多做过的题没写下来，如果我还能记得就补吧

## 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

5 rocks at distances 2, 11, 14, 17, and 21. Starting rock at position
0, finishing rock at position 25.

4

## HINT

移除之前，最短距离在位置2的石头和起点之间；移除位置2和位置14两个石头后，最短距离变成17212125之间的4

#include<cstdio>
#include<algorithm>
using namespace std;
int a[50010];
int L,n,m,l,r,ans;
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
inline bool jud(int x)
{
int now=0,kill=0,last=0;
while (++now<=n)
{
if (a[now]-a[last]<x)
{
kill++;
if (kill>m)return 0;
continue;
}else last=now;
}
return 1;
}
int main()
{
sort(a+1,a+n+1);
a[0]=0;a[++n]=L;
l=1;r=L;
while (l<=r)
{
int mid=(l+r)>>1;
if (jud(mid)){ans=mid;l=mid+1;}
else r=mid-1;
}
printf("%d",ans);
}

posted on 2014-07-31 21:35  zhber  阅读(109)  评论(0编辑  收藏  举报