River Hopscotch

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 12303 Accepted: 5271

Description

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 ≤ L ≤ 1,000,000,000). Along the river between the starting and ending rocks, N (0 ≤ N ≤ 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L).

To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river.

Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 ≤ M ≤ N).

FJ wants to know exactly how much he can increase the shortest distance before he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks.

Input

Line 1: Three space-separated integers: L, N, and M
Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position.

Output

Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks

Sample Input
25 5 2
2
14
11
21
17

Sample Output
4

Hint

Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).

題意大致是除M顆石頭讓最小間距的值最大

我的程式碼

#include<cstdio>
#include<algorithm>
using namespace std;
int L,N,M;int a[5000000+5];
int check(int ans)
{
    int last=0,now,m=0;
    while(last<N-1)
    {
        if(ans+a[last]<=L)
            now=lower_bound(a,a+N,ans+a[last])-a;
        else 
            now=N;
        m+=now-last-1;
        last=now;
    }
    return m>M;//注意 m==M 時 ans不一定是答案保留low=ans m>M時ans一定不是最終答案保留top=ans
}
int main()
{
    while(~scanf("%d%d%d",&L,&N,&M))
    {
        for(int i=0;i<N;i++)
            scanf("%d",&a[i]);
        if(N==M)
        {
            printf("%d\n",L);
            continue;
        }
        a[N++]=0,a[N++]=L;
        sort(a,a+N);
        int low=0,top=L,ans;
        while(top-low>1)//目的時找到使chech成立的最大答案 
        {
            int m=(top+low)>>1;
            int k=check(m);
            if(k)
                top=m;
            else
                low=m;
        }
        printf("%d\n",low);//low 是一定滿足條件的答案  top是一定不滿足條件的答案  滿足條件的最大值就是low 
    }
    return 0;
}

附上學長的程式碼留作參考

#include <cstdio>  
#include <algorithm>  
using namespace std;  
int w,n,k;  
int num[50000+11];  
bool check(int x)  
{  
    int pos,t;  
    int rec = k;  
    pos = lower_bound (num , num + n , x) - num;  
    rec = rec - pos;  
    if (rec < 0)  
        return false;  
    else if (pos == n)  
        return true;  
    while (rec >= 0)  
    {  
        t = lower_bound (num , num + n , num[pos] + x) - num;  
        rec = rec - (t - pos - 1);  
        if (rec < 0)  
            return false;  
        else if (t == n)  
            return true;  
        pos = t;  
    }  
}  
int main()  
{  
    int l,r,mid;  
    while (~scanf ("%d %d %d",&w,&n,&k))  
    {  
        num[0] = w;  
        for (int i = 1 ; i <= n ; i++)  
            scanf ("%d",&num[i]);  
        n++;  
        sort (num , num + n);  
        l = 1;  
        r = num[n-1];  
        while (r >= l)  
        {  
            mid = (l + r) >> 1;  
            if (check (mid))  
                l = mid + 1;  
            else  
                r = mid - 1;  
        }  
//      printf ("%d\n",l);      //這道題的情況特殊，結果並不是l  
        printf ("%d\n",l-1);  
    }  
    return 0;  
}