第十四届华中科技大学程序设计竞赛 K-Walking in the Forest 【二分】

最新推荐文章于 2018-07-12 18:03:24 发布

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本文介绍了一种解决特定路径寻找问题的方法，即在限定步数内找到从起点到终点跨越森林的最佳路径，使得单次跳跃的最大距离最小化。通过使用二分搜索算法，有效地解决了这一问题。

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

大写的门

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int MAX = 1e5 + 7;
ll a[MAX];
ll s, mx = 0, sum = 0, n, k;
void Bsearch()
{
    ll l = mx, r = sum;
    ll ans;
    while(l <= r)
    {
        ll mid = (l + r) >> 1;
        s = 0;
        ll S = 0;
        for(int i = 1; i <= n - 1; i++)
        {
            s += a[i];
            if(s > mid)
            {
                s = a[i];
                S++;
            }
        }
        if(S >= k)
            l = mid + 1;
        else
        {
            r = mid - 1;
            ans = mid;
        }
    }
    printf("%lld\n", ans);
}
int main()
{
    scanf("%lld%lld", &n, &k);
    for(int i = 1; i <= n - 1; i++)
    {
        scanf("%lld", &a[i]);
        mx = max(mx, a[i]);
        sum += a[i];
    }
    Bsearch();
    return 0;
}