uva 10382 - Watering Grass（贪心）

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针对UVA在线评测系统中的一道经典区间覆盖问题——浇水草地问题，通过贪心算法求解最小数量的喷水器来灌溉整个草地。输入包含喷水器的位置与作用范围，输出最少需要激活的喷水器数目。

题目链接http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1323

Problem E
Watering Grass
Input: standard input
Output: standard output
Time Limit: 3 seconds

n sprinklers are installed in a horizontal strip of grass l meters long andw 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 numbersn,l andw with n <= 10000. The next n lines 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

面积覆盖转化成经典的区间覆盖问题，贪心

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

struct Node{
    double s,e;
}a[10005];

bool cmp(Node x, Node y){
    return x.s < y.s;
}

int main(){
    int n,l,w,ans;
    int p,r;
    int i,j;
    double pos;//实时起点

    while(scanf("%d%d%d",&n,&l,&w) != EOF){
        ans = 0;
        for(i = 0;i < n;i++){
            cin >> p >> r;
            //除掉完全没用的
            if(r < w / 2.0){
                a[i].s = l+100;
                a[i].e = l+100;
                continue;
            }
            double tem = sqrt((double)r * r - (double)w * w  / 4.0);
            a[i].e = p + tem;
            a[i].s = p - tem;
        }
        sort(a,a+n,cmp);
        double max;
        int tag = 1;
        int last = 0;pos = 0;
        while(pos < l ){
            max = -1;
            for(i = last;(a[i].s < pos || abs(a[i].s - pos) < 1e-6)&& i < n;i++){
                if(a[i].e > max){
                    max = a[i].e;
                }
            }
            if(max == -1){
                tag = 0;
                break;
            }
            pos = max;
            last = i;
            ans++;
        }
        if(tag == 0)
            cout << -1 << endl;
        else
            cout << ans << endl;
    }

    return 0;
}

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