Divide and conquer:Aggressive Cows(POJ 2456)

　　　　　　　　　　　　　　　　侵略性的牛

　　题目大意：C头牛最大化他们的最短距离

　　常规题，二分法即可

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <functional>
 4
 5 using namespace std;
 6
 7 static int pos[100000];
 8
 9 bool judge(const int, const int,const int);
10 void solve(const int, const int);
11
12 int main(void)
13 {
14     int n, c;
15     while (~scanf("%d%d", &n, &c))
16     {
17         for (int i = 0; i < n; i++)
18             scanf("%d", &pos[i]);
19         sort(pos, pos + n);
20         solve(n, c);
21     }
22     return 0;
23 }
24
25 void solve(const int n, const int c)
26 {
27     int lb = 0, rb = 1000000001, mid;
28
29     while (rb - lb > 1)
30     {
31         mid = (rb + lb) / 2;
32         if (judge(mid, c, n)) lb = mid;
33         else rb = mid;
34     }
35     printf("%d\n", lb);
36 }
37
38 bool judge(const int dist, const int cows_sum, const int n)
39 {
40     int last_pos = 0, cnt;
41
42     for (int i = 1; i < cows_sum; i++)
43     {
44         cnt = last_pos + 1;
45         while (cnt < n && pos[cnt] - pos[last_pos] < dist)
46             cnt++;
47         if (cnt == n)
48             return false;
49         last_pos = cnt;
50     }
51     return true;
52 }