POJ 3258 River Hopscotch

River Hopscotch

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 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).

Source

USACO 2006 December Silver

数据范围很大，扫描会T，于是考虑分治。

拿走一块石头，该石头前面和后面的边会合并，由于要考虑新边的长度，所以必须保留石头之间的原顺序，那么就不能排序后分治边长来锁定答案了。

那么只好直接在0和最大长度之间二分确定答案。

每次按照尝试的答案扫描边，将长度小于答案的边全部删掉，记录拿走的石子个数。扫完之后，如果拿走的石子个数比限制多，说明答案不可行，继续二分；如果拿走的石子个数比限制少，说明还有更优解，继续二分。

 1 #include<algorithm>
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 using namespace std;
 7 const int mxn=100500;
 8 long long Le,n,m;
 9 long long dis[mxn];
10 int mid;
11 long long ans=0;
12 int cl(){
13     int cnt=0;
14     int lasum=0;
15     for(int i=1;i<=n;i++){
16         lasum+=dis[i]-dis[i-1];
17         if(lasum<mid){
18             cnt++;
19         }else lasum=0;//清零
20     }
21     if(cnt>m)return 0;//需要移除的石头数多于可移除石头数
22     return 1;
23 }
24 int main(){
25     scanf("%lld%lld%lld",&Le,&n,&m);
26     int i,j;
27     for(i=1;i<=n;i++)scanf("%lld",&dis[i]);
28     dis[n+1]=Le;
29     sort(dis+1,dis+n+2);
30     n+=1;
31     int l=0,r=Le;
32     while(l<=r){
33         mid=(l+r)>>1;
34         if(cl())//可行
35         {l=mid+1; ans=mid;}
36         else r=mid-1;
37     }
38     printf("%lld\n",ans);
39     return 0;
40 }