poj 3258(二分)

题意：有牛要过河，河宽为l，上面有n块石头，要求拿走m块石头后，让相邻两个石头的最小距离最大(假设两岸也有石头但无法取走)。
题解：二分出距离然后拿去判断。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 50005;
int l, n, m, pos[N];

bool judge(int x) {
    int cnt = 0, cur = 0;
    for (int i = 1; i <= n; i++) {
        while (i <= n && pos[i] - pos[cur] < x) {
            i++;
            if (cnt >= m)
                return false;
            cnt++;
        }
        cur = i;
    }
    if (l - pos[cur] < x)
        return false;
    return true;
}

int main() {
    while (scanf("%d%d%d", &l, &n, &m) == 3) {
        for (int i = 1; i <= n; i++)
            scanf("%d", &pos[i]);
        sort(pos + 1, pos + n + 1);
        pos[0] = 0;
        int left = 0, right = l;
        while (left < right) {
            int mid = (left + right + 1) / 2;
            if (judge(mid))
                left = mid;
            else
                right = mid - 1;
        }
        printf("%d\n", left);
    }
    return 0;
}