UVaOJ10382 - Watering Grass

最新推荐文章于 2019-07-21 17:47:34 发布

c随遇er安

最新推荐文章于 2019-07-21 17:47:34 发布

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分类专栏： Volume3-高效算法设计 ===ACM|ICPC=== 文章标签： ACM 算法 CC++

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本文介绍了一种算法解决浇水草地问题：通过计算最少喷头数确保整块草地得到灌溉。输入包括喷头数量、草地尺寸及各喷头位置与作用半径。采用排序和贪心策略，输出所需最小喷头数。

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10382 - Watering Grass

Time limit: 3.000 seconds

n sprinklers are installed in a horizontal strip of grass l meters long and w meters wide. Each sprinkler is installed at the horizontal center line of the strip. For each sprinkler we are given its position as the distance from the left end of the center line and its radius of operation.

What is the minimum number of sprinklers to turn on in order to water the entire strip of grass?

Input

Input consists of a number of cases. The first line for each case contains integer numbers n, l and w with n <= 10000. The next nlines contain two integers giving the position of a sprinkler and its radius of operation. (The picture above illustrates the first case from the sample input.)

Output

For each test case output the minimum number of sprinklers needed to water the entire strip of grass. If it is impossible to water the entire strip output -1.

Sample input

8 20 2

5 3

4 1

1 2

7 2

10 2

13 3

16 2

19 4

3 10 1

3 5

9 3

6 1

3 10 1

5 3

1 1

9 1

Sample Output

-1

(Regionals 2002 Warm-up Contest, Problem setter: Piotr Rudnicku)

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
using namespace std;

const int MAXN = 10005;
int n, nIndex;
double l, w;
struct Node {
    double left, right;
    friend bool operator < (const Node &a, const Node &b) {
        return a.left < b.left;
    }
}arr[MAXN];
int main() {
    double p, r;
    while (scanf("%d%lf%lf", &n, &l, &w) != EOF) {
        nIndex = 0;
        for (int i = 0; i < n; i ++) {
            scanf("%lf%lf", &p, &r);
            if (r < w / 2) continue;
            double t = sqrt(r * r - w * w / 4.0);
            arr[nIndex].left = p - t;
            arr[nIndex].right = p + t;
            nIndex ++;
        }
        sort(arr, arr + nIndex);
        int cnt = 0;
        double left = 0, right = 0;
        bool flag = false;
        if (arr[0].left <= 0) {
            int i = 0;
            while (i < nIndex) {
                int j = i;
                while (j < nIndex && left >= arr[j].left) {
                    if (arr[j].right > right) {
                        right = arr[j].right;
                    }
                    j ++;
                }
                if (j == i) break;
                cnt ++;
                i = j;
                left = right;
                if (left >= l) {
                    flag = true;
                    break;
                }
            }
        }
        if (flag) {
            printf("%d\n", cnt);
        } else {
            printf("-1\n");
        }
    }
    return 0;
}