The Frog's Games HDU - 4004 (二分)

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).

InputThe 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. OutputFor 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

每次做這種題都蒙圈，不過通過這幾種的二分，自己終於抓住了這種問題 的本質，但願以後不會卡這種題。

題意：青蛙跳河，河的寬度爲L， 裏面有n個石頭，青蛙可以跳到石頭上通過這些石頭過河，青蛙最多可以跳m次，求青蛙最大彈跳能力的最小值。

思路： 貪心加二分，要求青蛙最大彈跳能力的最小值，那麼我們就讓青蛙儘量跳m次，並且每次都盡全力跳，

可以通過二分青蛙的彈跳能力來找到解.

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int inf = 1000000100;
int a[500100];
int n,l,m;
bool ok(int x)
{
	int fr = 0;
	int step = 0;
	for(int i=1; i<=n+1;)
	{
	   if(fr >= l)break; // 一旦到了河對面就沒必要再跳了，感覺這邊也是個坑 
	   int t = i;
	   while(a[i]-fr <= x && i<=n+1){t=i;++i;}
       fr = a[t];
	   step++;
	   i = t;
	   if(step > m) return false;		
	}
    return true;		
}
int main()
{
	while(scanf("%d %d %d",&l, &n,&m) != EOF)
    {
		a[0] = 0;
		for(int i=1; i<=n; i++)
		{
		   scanf("%d", &a[i]);
		}
		a[n+1] = l;
		sort(a+1,a+1+n+1);
        int ll = 0;
		int r =inf;
		int ans = 0;
		while(ll<=r)
		{
		  int mid = (ll+r)/2;
		  if(ok(mid)){ans=mid; r=mid-1;}
		  else ll = mid+1;
		}       		
	   printf("%d\n",ans);	
	}
	return 0;
}

水波.