第十四届华中科技大学程序设计竞赛 KWalking in the Forest二分答案/最小化最大值

Posted 2020-11-02 Roni

链接：https://www.nowcoder.com/acm/contest/106/K
来源：牛客网

题目描述 
It’s universally acknowledged that there’re innumerable trees in the campus of HUST.

Now you're going to walk through a large forest. There is a path consisting of N stones winding its way to the other side of the forest. Between every two stones there is a distance. Let di indicates the distance between the stone i and i+1.Initially you stand at the first stone, and your target is the N-th stone. You must stand in a stone all the time, and you?can?stride over arbitrary number of stones in one step. If you stepped from the stone i to the stone j, you?stride a span of?(di+di+1+...+dj-1). But there is a limitation. You're so tired that you want to walk through the forest in no more than K steps. And to walk more comfortably, you have to minimize the distance of largest step.
输入描述:
The first line contains two integer N and K as described above.
Then the next line N-1 positive integer followed, indicating the distance between two adjacent stone.
输出描述:
An integer, the minimum distance of the largest step.
示例1
输入
6 3
1 3 2 2 5
输出
5

【题意】：
题意就是说有n块石头，每块石头中间有一定的距离，一次可以跳过多个石头，但是不可以超过k步，求最大步的最小值 。

其实意思就是说，如果你每一步比较小，比如一块一块石头地过去，那么步数就太多了；但是如果你直接一步跳到最后，这样又太浪费体力了，而且不符合题意，而题意就是要你找到这么一个平衡点，可以恰好走到三步，然后求其中最大的一步的距离（相对于其他两步是最大的，但相对于所有情况的最大步它是最小的一种情况）

【出处】：POJ 3272 Monthly Expense

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <set>
#include <string>
#include <cstdlib>
#include <cmath>
using namespace std;
#define ll long long
#define mod 1000000007
int n, k;
ll a[100005];
int main() {
    scanf("%d%d", &n, &k);
        ll s = 0;
        ll Max = 0;
        for (int i = 1; i <= n - 1; ++i) {
            scanf("%lld", &a[i]);
            s += a[i];
            Max = max(Max, a[i]);
        }
        ll l = Max, r = s, ans, mid;
        while (l <= r) {
            mid = (l + r) >> 1;
            s = 0;
            int c = 0;
            for (int i = 1; i <= n - 1; ++i) {
                s += a[i];
                if (s > mid) {   //多个跳不过，只能前面算跳一次，从这里重新开始跳
                    s = a[i];
                    c++;
                }
            }

            if (c >= k) {
                l = mid + 1;
            }
            else {
                r = mid - 1;
                ans = mid;
            }
        }
        cout << ans << endl;
    return 0;
}