River Hopscotch解题报告

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 toMrocks (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, andM
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 2214112117

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).

 

数轴上有n个石子，第i个石头的坐标为Di，现在要从0跳到L，每次条都从一个石子跳到相邻的下一个石子。
现在FJ允许你移走M个石子，问移走这M个石子后，相邻两个石子距离的最小值的最大值是多少。

#include<iostream> #include<algorithm> using namespace std; int main(void) {     int L,n,m;        while(cin>>L>>n>>m)     {         int* dist=new int[n+2];  //新的定义数组的方法。         dist[0]=0;             dist[n+1]=L;           int low=L,high=L;                   for(int i=1;i<=n+1;i++)         {             if(i<=n)     //n+1不用输入。                 cin>>dist[i];             if(low > dist[i]-dist[i-1]) //找出最小差值。                 low=dist[i]-dist[i-1];         }         sort(dist,dist+(n+2));     //排序。              while(low<=high)    //查找的x在（low，high）中，不断逼近。         {             int mid=(low+high)/2,delrock=0,sum=0;               for(int i=1;i<=n+1;)             {                 if( (sum+=(dist[i]-dist[i-1]))<=mid) //可能两石头间（《mid。且有几块石头）可移动sum个都满足。                 {                     i++;                     delrock++;  //移动一块石头++；                 }                 else   //不满足就是起点往后移。                 {                     i++;                     sum=0;                   }             }             if(delrock<=m)    //移动的石头<m，增大mid.                 low=mid+1;                else             //减少mid。                 high=mid-1;           }              cout<<low<<endl;      delete dist;  //释放dist。     }     return 0; }