The Frog's Games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 3263    Accepted Submission(s): 1596

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

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

 參考別人的思路做的，挫....

利用二分來選擇適合的數。。。然後再驗證，是列舉的優化版。 好吧，這個思路是別人的 ，我還是太挫啦

程式碼：

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<functional>
 5 const int maxn= 500005;
 6 using namespace std;
 7 int stone[maxn],n,m;
 8 bool judge(int dis)  
 9 {
10     int i=1,pre=0,count=0;
11     while(i<=n+1)
12     {
13         count++;
14         if(dis<stone[i]-stone[i-1])  return false ;  //這個石頭跳不過去，所以失敗
15         while (i<=(n+1)&&dis>=stone[i]-stone[pre]) i++;
16         pre=i-1;
17         if(count>m)   return false ;
18     }
19     return true ;
20 }
21 int main()
22 {
23     int length,i;
24     while(scanf("%d%d%d",&length,&n,&m)!=EOF)
25     {
26          memset(stone,0,sizeof(int)*(n+3));
27           for(i=1;i<=n;i++)
28          scanf("%d",&stone[i]);
29           stone[i]=length;
30        sort(stone,stone+(n+2),less<int>());   //升序安放stone
31        int ans,low=0,high=length;
32         while(low<=high)
33         {
34              ans=(low+high)/2;   //跳這麼遠的時候能否滿足要求
35             if(ans*m>=length&&judge(ans))  high=ans-1;
36             else low=ans+1;
37         }
38         printf("%dn",low);
39     }
40     return 0;
41 }