HDU-4004 The Frog's Games （分治）

http://acm.hdu.edu.cn/showproblem.php?pid=4004

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

Sample Output

题意：

青蛙王国运动会开始了，最受欢迎的游戏是铁蛙三项赛，其中一项是跳跃过河项目。这个项目需要青蛙运动员通过跳跃过河。河的宽度是L。在河面上有直线排列的n个石头。青蛙可以利用这些石头跳跃过河，如果落入河中则失败。青蛙们能够跳跃的最多次数是m。现在想要知道青蛙们至少需要具备多大的跳跃距离，才能够顺利完成比赛。

思路：

用二分去查找一个单次跳跃的最大距离，看以这个距离能否完成。直到找到最小的那个数，二分的上边界是河水宽 L。

代码如下：

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <string>
 #include <math.h>
 #include <algorithm>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #include <sstream>
 const int INF=0x3f3f3f3f;
 typedef long long LL;
 const int mod=1e9+;
 //const double PI=acos(-1);
 #define Bug cout<<"---------------------"<<endl
 const int maxn=5e5+;
 using namespace std;

 int a[maxn];//存放每块石头到开始处的距离
 int b[maxn];//存放石头之间的差值 

 int main()
 {
     int L,n,m;
     while(~scanf("%d %d %d",&L,&n,&m))
     {
         for(int i=;i<=n;i++)
             scanf("%d",&a[i]);
         a[n+]=L;//最后要跳到河岸上
         sort(a+,a++n+);
         int MAX=;//答案不可能会小于石头差值中的最大值，否则这两块石头一定跳不过去
         for(int i=;i<=n+;i++)
         {
             b[i]=a[i]-a[i-];
             if(b[i]>MAX)
                 MAX=b[i];
         }
         int l=MAX;
         int r=L;
         while(l<=r)//二分查找答案
         {
             int mid=(l+r)>>;//二分法的中值（即初始青蛙能跳的最大距离）
             int num=;//记录跳的步数
             for(int i=;i<=n+;)//此处用到贪心，一次尽量多跳几个石头
             {
                 int sum=;//检验到这个石头需要跳的距离
                 for(int j=i;j<=n+;j++)
                 {
                     if(sum+b[j]<=mid)//可以继续跳
                     {
                         sum+=b[j];
                         if(j==n+)//跳到岸了，处理一下
                         {
                             num++;
                             i=n+;
                             break;
                         }
                     }
                     else//不可以继续跳
                     {
                         num++;//跳的次数加1
                         i=j;//下一次从这个石头起跳
                         break;
                     }
                 }
             }
             if(num>m)
                 l=mid+;
             else
                 r=mid-;
         }
         printf("%d\n",l);
     }
     return ;
 }