Day3-R-Aggressive cows POJ2456

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：可验证，单调，二分答案，代码如下：

（左闭右开）

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

（左闭右闭）

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

小结：对于问题可进行贪心验证+答案具有单调性可以进行二分，注意左闭右闭的while条件