uva-10382-Watering Grass

A - Watering Grass

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit  Status

Description

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 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 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

6 

2

-1

转换成最小区间覆盖

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>


using namespace std;


#define F(i, n) for (int i=0;i<(n);++i)
#define LBIT(x) x&-x


int T;
long long a, b, c, d, p, m;


long long gcd(long long a, long long b) {
    return b == 0 ? a : gcd(b, a % b);
}


template<class T>
void myswap(T * a, T * b) {
    T tmp = *a;
    *a = *b;
    *b = tmp;
}


#define MAXSIZE 10000
double ml[MAXSIZE + 1];
double mr[MAXSIZE + 1];
vector<int> arr;


bool comp(int a, int b) {
    if (ml[a] < ml[b])
        return true;


    return false;
}


int main()
{
    freopen("input.in", "r", stdin);
    int a;
   // cin >> a;
    int n, l, w;
    while(scanf("%d%d%d", &n, &l, &w) == 3) {
        int _min, _max;
        double pos, r;
        double _l, _r;
        int cnt = 0;
        int ans = 0;
        int index;
        double w2 = w*w*1.0/4;


        arr.clear();
        F (i, n) {
            cin >> pos >> r;
            _l = pos - sqrt(r * r - w2);
            _r = pos  + sqrt(r * r - w2);


            if (r * 2 >= w) {
                ml[cnt] = _l;
                mr[cnt] = _r;
                arr.push_back(cnt++);
            }
        }


        if (cnt == 0) {
            if (l > 0)
                cout << -1 << endl;
            else
                cout << 0 << endl;
        }


        sort(arr.begin(), arr.end(), comp);


        _l = ml[arr[0]];
        _r = mr[arr[0]];
        index = 1;
        ans = 1;


        if (_l > 0) {
            cout << -1 << endl;
            continue;
        }


        while(index < cnt && _r < l) {
            r = -1;


            if (ml[arr[index]] > _r)
                break;


            while(index < cnt && ml[arr[index]] <= _r) {
                if (mr[arr[index]] > r)
                    r = mr[arr[index]];
                index++;
            }


            _r = r;
            ans++;
        }


        if (_r >= l)
            cout << ans << endl;
        else
            cout << -1 << endl;


    }


    return 0;
}