poj2456 Aggressive cows（二分查找）

https://vjudge.net/problem/POJ-2456

二分，从最大长度开始，不断折半试，如果牛全放下了，就是可行，修改下界，否则改上届。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<map>
 8 #define lson l, m, rt<<1
 9 #define rson m+1, r, rt<<1|1
10 #define INF 0x3f3f3f3f
11 typedef unsigned long long ll;
12 using namespace std;
13 int n, m, a[100010];
14 int C(int x)
15 {
16     int pre=0, ans=1;
17     for(int i = 1; i < n; i++){
18         while(i < n&&a[i]-a[pre]<x){
19             i++;
20         }
21         if(i>=n) break;
22         ans++;
23         pre = i;
24     }
25     return ans >= m;
26 }
27 int main()
28 {
29     while(~scanf("%d%d", &n, &m)){
30         for(int i = 0; i < n; i++){
31             scanf("%d", &a[i]);
32         }
33         sort(a, a+n);
34         int lb = 0, ub = INF;
35         while(ub-lb>1){
36             int mid = (ub+lb)/2;
37             if(C(mid)){
38                 lb = mid;
39             }
40             else ub = mid;
41         }
42         printf("%d\n", lb);
43     }
44     return 0;
45 }