The Frog's Games

The Frog's Games

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

分析：

就是一个青蛙要跳的对岸，然后河流中间有M块石头，然后最多可以跳k次，问这个青蛙最短需要跳多远才可以保证不掉下去。按照我的想法，m个石头，都被跳到的话需要跳M+1次。那么如果k >= M+1次，那就是找到两块石头之间距离最大的哪一个就可以了。但是如果k < M+1次的话呢，青蛙就需要一次跳过多的石头，M+1 - k.就可以知道青蛙需要越过多少块石头，既在多少块石头上面不能停留。所以呢，我就需要每次去遍历，找到最近的相隔一块石头的两块石头。那么我就需要循环遍历(M+1-k) * (M+1)  次左右了。这肯定不行啊,大致估计一下时间复杂度500000 * 500000，GG，肯定会超时。想了一想，想不到好办法，然后百度，说用二分，去二分所要求的跳跃能力，然后每次检验一下能不能跳过去。这样再算一下时间，500000 * log(1000000000 ),这不会超时。应该记住的 n*2 的时间优化多半是优化到n*log(n), 而log(n)的优化，多半是二分，或者一些STL的数据结构。

 #include<bits/stdc++.h>

 using namespace std;

 const int N = ;

 int ans[N];

  int ll, nn, mm;

 bool judge (int dis) {
   int tx = , i, ty = ;
   for (int i = ; i <= mm; i++) {
     while (ty <= nn +  && ans[ty] - ans[tx] <= dis) ty++;

     ty --;
     tx = ty;
   }
   return ty == nn + ;
 }

 int main () {

   while (~scanf("%d %d %d", &ll, &nn, &mm)) {

     for (int i = ; i <= nn; i++) {
       scanf("%d", &ans[i]);
     }
     ans[] = ; ans[nn+] = ll;
     sort(ans, ans + nn +);
     int L = , R = ll;
     while (L <= R) {
       int mid = (L + R) >> ;
       if (judge(mid)) {
         R = mid - ;
       } else L = mid + ;
     }
     printf("%d\n", L);
   }
   return ;
 }

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