hdoj 4004 The Frog's Games

D - The Frog's Games

Time Limit:1000MS     Memory Limit:65768KB     64bit IO Format:%I64d & %I64u

Submit

 

Status

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 

二分。。。。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
LL l,n,m,le,ri,mid,ans;
int shu[600000];
LL ll[600000];
int zhao(LL xx)
{
	int kp=1;
	LL sum=0;
	for (int i=0;i<=n;i++)
	{
		if (sum+ll[i]>xx)
		{
			kp++;
			sum=ll[i];
		}
		else
		sum+=ll[i];
	}
	if (kp<=m)
	return true;
	else
	return false;
}
int main()
{
	while (~scanf("%lld%lld%lld",&l,&n,&m))
	{
		memset(shu,0,sizeof(shu));
		for (int i=1;i<=n;i++)
		scanf("%d",&shu[i]);
		sort(shu+1,shu+1+n);
		shu[0]=0;
		ri=0;le=0;
		for (int i=0;i<n;i++)
		{
			ll[i]=shu[i+1]-shu[i];
			le=max(le,ll[i]);
			ri+=ll[i];
		}
		ll[n]=l-shu[n];
		le=max(le,ll[n]);
		ri+=ll[n]; 
		while (ri>=le)
		{
			mid=(ri+le)/2;
			if (zhao(mid))
			{
				ans=mid;
				ri=mid-1;
			}
			else
			le=mid+1;
		}
		printf("%lld\n",ans);
	}
	return 0;
}