hdu 4004 The Frog's Games

The Frog's Games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1582    Accepted Submission(s): 837

Problem Description

The annual Games in frogs' kingdom started again. The most famous game is the Ironfrog Triathlon. One test in the Ironfrog Triathlon is jumping. This project requires the frog athletes to jump over the river. The width of the river is L (1<= L <= 1000000000). There are n (0<= n <= 500000) stones lined up in a straight line from one side to the other side of the river. The frogs can only jump through the river, but they can land on the stones. If they fall into the river, they
are out. The frogs was asked to jump at most m (1<= m <= n+1) times. Now the frogs want to know if they want to jump across the river, at least what ability should they have. (That is the frog's longest jump distance).

 

Input

The input contains several cases. The first line of each case contains three positive integer L, n, and m.
Then n lines follow. Each stands for the distance from the starting banks to the nth stone, two stone appear in one place is impossible.

 

Output

For each case, output a integer standing for the frog's ability at least they should have.

 

Sample Input

    
    
    
    

     
     
     
     6 1 2
    
    
    
    
    
    
    
    

     
     
     
     2
    
    
    
    
    
    
    
    

     
     
     
     25 3 3
    
    
    
    
    
    
    
    

     
     
     
     11 
    
    
    
    
    
    
    
    

     
     
     
     2
    
    
    
    
    
    
    
    

     
     
     
     18
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     4
    
    
    
    
    
    
    
    

     
     
     
     11
    
    
    
    
    
    
    
    

     
     
     
     
/*题意 一只青蛙过河 要跳 中间有n个石头 最多跳m次 河的长度为l
 每个方案中一步的大小为1到k  问多种方案中k最小的一个是？*/#include<stdio.h>
#include<stdlib.h>
int l,m,n,a[500000+5];
int cmp(const void *a,const void *b)
{
 return *(int *)a-*(int *)b;
}
int can_cross(int k)//步子大小为k
{
 int i,last_pos=0,step=0;
 if(k<a[1]) return 0;
 for(i=1;i<n;i++)
 {
  if(a[i+1]-a[i]>k) return 0;
  if(a[i+1]-last_pos<=k) continue;
  if(a[i]-last_pos<=k)  {step++;last_pos=a[i];}
 }
 if(a[n]-a[n-1]>k) return 0;
    else step++;
 if(step>m)
      return 0;
 else return 1;}
int main()
{
   int i,start,end,mid,ans;
   while(scanf("%d %d %d",&l,&n,&m)!=EOF)
   {
         for(i=1;i<=n;i++)
    scanf("%d",&a[i]);
   a[++n]=l;
   qsort(&a[1],n,sizeof(a[0]),cmp);
         start=l/m;end=l;
   while(start<=end)//这个地方是等于 不然就WA了
   {
    mid=(start+end)/2;
              if(can_cross(mid))
     {
      end=mid-1;
      ans=mid;
     }
     else start=mid+1;
   }
   printf("%d\n",ans);
   }
   return 0;
}