GZHU18級寒假訓練：Virgo's Trial-J

UVA-10382

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

不确定這題算不算DP，但是是要一個數字記錄被覆寫到的草地，若果所有圓都覆寫不了，就輸出不可能。

#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <string.h>
using namespace std;
 
const int N = 10005;
int n, l, w;
int visit[N];
 
double Max (const double a, const double b) { return a > b ? a: b; }
 
struct Len {
	
	double L, R;	
}len[N];
 
int cmp (const Len &a, const Len &b) { return a.L < b.L;}
 
void handle (int c, int r, int i) {
	
	double r1 = r;
	double c1 = c;
	double w1 = w;
	double temp = sqrt (r1 * r1 - 0.25 * w1 * w1);
	len[i].L = c - temp;
	len[i].R = c + temp;
}
 
int solve () {
	
	memset (visit, 0, sizeof (visit));
	if (len[0].L > 0)
		return -1;
	double s = 0;
	double ll = -1;
	int m = -1;
	for (int i = 0; i < n; i++) {
		
		if (len[i].L <= s) {
			
			if (ll < len[i].R) {
 
				visit[i] = 1;
				if (m >= 0)
					visit[m] = 0;
				m = i;
				ll = Max (ll, len[i].R);
			}
		} else {
			
			if (s == ll)
				return -1;
			s = ll;
			i--;
			m = -1;
		}
//		printf ("%f\n", ll);
	}
//	printf ("%f\n", ll);
	if (ll < l)
		return -1;
	int count = 0;
	for (int i = 0; i < n; i++)
		if (visit[i])
			count++;
	return count;
}
 
int main () {
	
	int c, r;
	while (scanf ("%d%d%d", &n, &l, &w) != EOF) {
 
		for (int i = 0; i < n; i++) {
 
			scanf ("%d%d", &c, &r);
			handle (c, r, i);
		}
		sort (len, len + n, cmp);
		
/*		for (int i = 0; i < n; i++)
			printf ("%f %f\n", len[i].L, len[i].R);*/
		printf ("%d\n", solve());
 
	}
	return 0;
}



           

代碼來源：https://blog.csdn.net/u012997373/article/details/38150585