Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000).
His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ want to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?
Input
* Line 1: Two space-separated integers: N and C
* Lines 2..N+1: Line i+1 contains an integer stall location, xi
Output
* Line 1: One integer: the largest minimum distance
Sample Input
5 3
1
2
8
4
9 Sample Output
3 Hint
OUTPUT DETAILS:
FJ can put his 3 cows in the stalls at positions 1, 4 and 8, resulting in a minimum distance of 3.
Huge input data,scanf is recommended. 思路：可驗證，單調，二分答案，代碼如下： （左閉右開）
const int maxm = 100010;
int buf[maxm], n, k;
bool check(int d) {
int t = 1, now = buf[0];
for (int i = 1; i < n; ++i) {
if(buf[i] - now >= d) {
now = buf[i];
t++;
}
if(t >= k)
return true;
}
return false;
}
int main() {
scanf("%d%d", &n, &k);
for (int i = 0; i < n; ++i)
scanf("%d", &buf[i]);
sort(buf, buf + n);
int l = 1, r = buf[n - 1] - buf[0] + 1, mid;
while(l < r) {
mid = (l + r) >> 1;
if(check(mid))
l = mid + 1;
else
r = mid;
}
printf("%d\n", l - 1);
return 0;
} View Code
（左閉右閉）
const int maxm = 100010;
int buf[maxm], n, k;
bool check(int d) {
int t = 1, now = buf[0];
for (int i = 1; i < n; ++i) {
if(buf[i] - now >= d) {
now = buf[i];
t++;
}
if(t >= k)
return true;
}
return false;
}
int main() {
scanf("%d%d", &n, &k);
for (int i = 0; i < n; ++i)
scanf("%d", &buf[i]);
sort(buf, buf + n);
int l = 1, r = buf[n - 1] - buf[0], mid;
while(l <= r) {
mid = (l + r) >> 1;
if(check(mid))
l = mid + 1;
else
r = mid - 1;
}
printf("%d\n", r);
return 0;
} View Code
小結：對于問題可進行貪心驗證+答案具有單調性可以進行二分，注意左閉右閉的while條件
轉載于:https://www.cnblogs.com/GRedComeT/p/11249568.html