杭电 4004 The Frog's Games 青蛙跳水 （二分法，贪心）

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

 

       输入河宽L，石头数量N，步数M，在区间【1，L】内用二分法判断，算出区间里数作为一步的最大值时到对岸（最优解）需要多少的步数，若步数大于等于m，则记下步数，令右区间减一，否则令左区间加一，直到求出最小的能力。

#include<cstdio>
#include<algorithm>
using namespace std;
int a[500000+11];
int l,n,m,i,le,ri,mid,ans;
bool f(int k)　　　　　　//返回值为真或假
{
    int num=0,sum=0;
    if(a[0] > k) 
        return 0;
    for(i = 1 ; i <= n ; i++)
    {
        if(a[i] - a[i-1] > k)
            return 0;
        if((a[i]-sum) > k)
        {
            num++;
            sum=a[i-1];
        }
    }
    return (num+1) <= m;
}
int main()
{    
    while(scanf("%d %d %d",&l,&n,&m)!=EOF)
    {
        for(i = 0 ; i < n ; i++)
         {
             scanf("%d",&a[i]);
         }
         a[n]=l;
         sort(a,a+n);
         le=1;
         ri=l;
         while(le <= ri)
         {
             mid=(le+ri)/2;
             if( f(mid))
             {
                 ans=mid;
                 ri=mid-1;
             }
             else
             {
                 le=mid+1;
             }
            }
        printf("%d
",ans);
    }
}

 

#include<cstdio>
#include<algorithm>
using namespace std;
int a[500000+11];
int l,n,m,i,le,ri,mid,ans;
int f(int k)　　　　　　//返回值为最大能力为k时的步数
{
    int num=0,sum=0;
    if(a[0] > k) 
        return m+1;
    for(i = 1 ; i <= n ; i++)
    {
        if(a[i] - a[i-1] > k)
            return m+1;
        if((a[i]-sum) > k)
        {
            num++;
            sum=a[i-1];
        }
    }
    return num+1;
}
int main()
{    
    while(scanf("%d %d %d",&l,&n,&m)!=EOF)
    {
        for(i = 0 ; i < n ; i++)
         {
             scanf("%d",&a[i]);
         }
         a[n]=l;
         sort(a,a+n);
         le=1;
         ri=l;
         while(le <= ri)
         {
             mid=(le+ri)/2;
             if( f(mid) <= m)
             {
                 ans=mid;
                 ri=mid-1;
             }
             else
             {
                 le=mid+1;
             }
            }
        printf("%d
",ans);
    }
}