「USACO 2012 March (Silver)」Flowerpot

#题面

#题目描述

Farmer John has been having trouble making his plants grow, and needs your help to water them properly. You are given the locations of $N$ raindrops ($1 \leq N \leq 10^5$) in the 2D plane, where $y$ represents vertical height of the drop, and $x$ represents its location over a 1D number line:

Each drop falls downward (towards the $x$-axis) at a rate of $1$ unit per second. You would like to place Farmer John’s flowerpot of width $W$ somewhere along the $x$ axis so that the difference in time between the first raindrop to hit the flowerpot and the last raindrop to hit the flowerpot is at least some amount $D$ (so that the flowers in the pot receive plenty of water). A drop of water that lands just on the edge of the flowerpot counts as hitting the flowerpot.

Given the value of $D$ and the locations of the $N$ raindrops, please compute the minimum possible value of $W$.

#输入格式

• Line $1$: Two space-separated integers, $N$ and $D$. ($1 \leq D \leq 10^6$)
• Lines $2 \sim N + 1$: Line $i + 1$ contains the space-separated $(x, y)$ coordinates of raindrop $i$, each value in the range $[0, 10^6]$.

#输出格式

• Line $1$: A single integer, giving the minimum possible width of the flowerpot. Output $-1$ if it is not possible to build a flowerpot wide enough to capture rain for at least $D$ units of time.

#输入输出样例

样例输入 #1

4 5
6 3
2 4
4 10
12 15

样例输出 #1

2

样例解释 #1

There are $4$ raindrops, at $(6, 3)$, $(2, 4)$, $(4, 10)$, and $(12, 15)$. Rain must fall on the flowerpot for at least $5$ units of time.

A flowerpot of width $2$ is necessary and sufficient, since if we place it from $x = 4 \dots 6$, then it captures raindrops $1$ and $3$, for a total rain duration of $10-3 = 7$.

#思路

考场上想出来的做法，题解里好像没有一样的思路。

本题需要注意的点是，第一滴水和最后一滴水之间间隔的时间不一定恰好为 $D$，而是 $\geq D$，一定要看清楚题，场上因为这个浪费了一个小时的时间。

将所有点按 $y$ 值排序之后，开一个 set 记录 $y_j \leq y_i - d$ 的对应的 $x_j$ 值。因为题目中没有对第一滴水和最后一滴水之间接水数量之间做要求，所以直接在 set 中查询 $x_i$ 的前驱和后继，取两个方向上长度的最小值更新答案即可。

#代码

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#include <iostream>
#include <algorithm>
#include <cmath>
#include <limits>
#include <set>
#include <utility>

using std::cin;
using std::cout;
const char endl = '\n';

const int N = 1e5 + 5;

int n, d, ans = std::numeric_limits<int>::max();
std::pair<int, int> points[N];
std::set<int> set;

int main() {
    std::ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> n >> d;

    for (int i = 1; i <= n; i++) {
        cin >> points[i].first >> points[i].second;
    }

    std::sort(points + 1, points + 1 + n, [](auto a, auto b) -> bool {
        return a.second < b.second;
    });

    int lst = 1;
    for (int i = 1; i <= n; i++) {
        int pos = std::upper_bound(
                      points + 1,
                      points + 1 + n,
                      std::make_pair(0, points[i].second - d),
                      [](auto a, auto b) -> bool {
                          return a.second < b.second;
                      })
                - points;

        for (int j = lst; j < pos; j++) {
            set.insert(points[j].first);
        }

        lst = pos;

        auto it1 = set.lower_bound(points[i].first),
             it2 = set.upper_bound(points[i].first);

        if (it1 != set.begin()) {
            ans = std::min(ans, std::abs(points[i].first - *--it1));
        }

        if (it2 != set.end()) {
            ans = std::min(ans, std::abs(points[i].first - *it2));
        }
    }

    cout << (ans == std::numeric_limits<int>::max() ? -1 : ans) << endl;

    return 0;
}