POJ 3258 River Hopscotch binary answer

River Hopscotch

Time Limit:2000 MS		Memory Limit:65536 K
Total Submissions:6193		Accepted:2685

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,LUnits 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 distanceDiFrom 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 compression ss 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 upMRocks (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 setMRocks.

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 MRocks

Sample Input

25 5 2214112117

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 ).

Source

USACO 2006 December Silver
Question

It is also a question of the second answer. The question is: a river with a length of L has N stepping stones. Now we need to remove m stepping stones, this maximizes the distance between two stones. The cross-Strait of the river can also be seen as a stepping stone, but they cannot be removed.

Just like the previous questions, you can use the bipartite answer range. However, this question differs from the previous ones in that it updates res when modifying the search upper bound (the previous questions are updated when the lower bound is modified ). Return res.

The result is exactly the one in the sample code.

Sample Code

/*****@author    Shen*@title     poj 1064*/#include 
 
  #include 
  
   #include 
   
    #include using namespace std;int n, k;int l, v[50005];int maxa = 0;bool test(int x){    int sum = 0, st = 0;    for (int i = 1; i <= n + 1; i++)    {        if (v[i] - v[st] <= x)            sum++;        else            st = i;    }    //printf("\t%s with x = %d, result is that sum = %d.\n", __func__, x, sum);    return sum <= k;}int Bsearch(int l, int r){    int res = r;    while (l <= r)    {        int mid = (r + l) / 2;        //printf("l = %d, r = %d, mid = %d, res = %d.\n", l, r, mid, res);        if (test(mid))             l = mid + 1;        else            res = min(res, mid), r = mid - 1;    }    return res;}void solve(){    maxa = l;    v[0] = 0;    for (int i = 1; i <= n; i++)    {        scanf("%d", &l);        v[i] = l;    }    v[n + 1] = maxa;    sort(v, v + n + 2);    int ans = Bsearch(0, maxa);    printf("%d\n", ans);}int main(){    while (~scanf("%d%d%d",&l, &n, &k))        solve();    return 0;}