HDU Problem 4004 The Frog's Games 【二分】

The Frog's Games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 6176    Accepted Submission(s): 3010

Problem Description

The annual Games in frogs' kingdom started again. The most famous game is the Ironfrog Triathlon. One test in the Ironfrog Triathlon is jumping. This project requires the frog athletes to jump over the river. The width of the river is L (1<= L <= 1000000000). There are n (0<= n <= 500000) stones lined up in a straight line from one side to the other side of the river. The frogs can only jump through the river, but they can land on the stones. If they fall into the river, they
are out. The frogs was asked to jump at most m (1<= m <= n+1) times. Now the frogs want to know if they want to jump across the river, at least what ability should they have. (That is the frog's longest jump distance).

 

 

Input

The input contains several cases. The first line of each case contains three positive integer L, n, and m.
Then n lines follow. Each stands for the distance from the starting banks to the nth stone, two stone appear in one place is impossible.

 

 

Output

For each case, output a integer standing for the frog's ability at least they should have.

 

 

Sample Input

6 1 2 2 25 3 3 11 2 18

 

 

Sample Output

4 11

 

青蛙想跳过长为L的河面，河上有n块石头，他们想只跳过m个石头。利用二分吗，寻找最优解。

#include <cstdio>
#include <cmath>
#include <algorithm>
#define MAX_N 500050
using namespace std;
const double ESP = 1e-5;
const int INF = 1e8;

int stone[MAX_N];
int L, m, n;

bool judge(int x) {
    int last = 0;
    int cnt = 0;
    for (int  i = 0; i <= n; ) {
        if (last + x >= stone[i]) {
            i++;
        }
        else {
            if (last == stone[i - 1] || i == 0)
                return false;
            last = stone[i - 1];
            cnt++;
        }
    }
    cnt++;
    return cnt <= m;
}

int main() {
    while (scanf("%d%d%d", &L, &n, &m) != EOF) {
        for (int i = 0; i < n; i++) {
            scanf("%d", &stone[i]);
        }
        sort(stone, stone + n);
        int l = 0, r = L, mid, ans;
        stone[n] = L;
        for (int i = 0; i < 100; i++) {
            mid = (l + r)/2;
            if (judge(mid)) {
                ans = mid;
                r = mid - 1;
            }
            else  l = mid + 1;
        }
        printf("%d\n", ans);
    }
    return 0;
}