HDU4004 The Frog's Games

HDU4004 The Frog's Games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)

Total Submission(s): 941    Accepted Submission(s): 508

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

******************************************************************

这是2011年大连赛区网络赛的第四题，简单二分答案+贪心，不解释。

#include <stdio.h>
#include <algorithm>

int pos[500005];
int len,n,tim;
int max(int a,int b){return a>b?a:b;}

int ifok(int k)
{
    int now=0;
    for(int i=1;i<=tim;i++)
    {
        int j;
        for(j=now;pos[j+1]-pos[now]<=k;j++);
        now=j;
        if(now==n+1)return 1;
    }
    return 0;
}

int main()
{
    while(scanf("%d%d%d",&len,&n,&tim)==3)
    {
        int maxx=-1;
        pos[0]=0;
        for(int i=1;i<=n;i++)
            scanf("%d",&pos[i]);
        std::sort(pos+1,pos+1+n);
        for(int i=1;i<=n;i++)
            maxx=max(maxx,pos[i]-pos[i-1]);
        pos[n+1]=len;
        pos[n+2]=0x3f3f3f3f;
        maxx=max(maxx,pos[n+1]-pos[n]);
        int le=maxx-1,ri=len,mid;
        while(mid=(ri+le)/2,ri>le)
        {
            if(ifok(mid))
                ri=mid;
            else
                le=mid+1;
        }
        printf("%d\n",ri);
    }
}