O - Muddy roads

Farmer John has a problem: the dirt road from his farm to town has suffered in the recent rainstorms and now contains (1 <= N <= 10,000) mud pools.

Farmer John has a collection of wooden planks of length L that he can use to bridge these mud pools. He can overlap planks and the ends do not need to be anchored on the ground. However, he must cover each pool completely.

Given the mud pools, help FJ figure out the minimum number of planks he needs in order to completely cover all the mud pools.

Input

* Line 1: Two space-separated integers: N and L

* Lines 2..N+1: Line i+1 contains two space-separated integers: s_i and e_i (0 <= s_i < e_i <= 1,000,000,000) that specify the start and end points of a mud pool along the road. The mud pools will not overlap. These numbers specify points, so a mud pool from 35 to 39 can be covered by a single board of length 4. Mud pools at (3,6) and (6,9) are not considered to overlap.

Output

* Line 1: The miminum number of planks FJ needs to use.

Sample

Inputcopy	Outputcopy
3 3 1 6 13 17 8 12	5

Hint

INPUT DETAILS:

FJ needs to use planks of length 3 to cover 3 mud pools. The mud pools cover regions 1 to 6, 8 to 12, and 13 to 17.

OUTPUT DETAILS:

FJ can cover the mud pools with five planks of length 3 in the following way:

                   111222..333444555....

                   .MMMMM..MMMM.MMMM....

                   012345678901234567890

题义：
农民约翰想用几块等长的木板遮住他的农场，输入遮住距离有几段和木板的长度，和每段被遮住位置的起止位置，求出所用木板最少的数量。

#include
#include
#include
using namespace std;
typedef long long ll;

const int N = 1e5 + 10;

//建立结构体记录泥坑
struct node{
	int l;
	int r;
	bool operator < (struct node& a) { return l < a.l; } //重载运算符
}a[N];

int main()
{
	ios::sync_with_stdio(false);

	int n, m; cin >> n >> m;

	for (int i = 1; i <= n; i++) cin >> a[i].l >> a[i].r;

	sort(a + 1, a + n + 1); //对泥坑的起始点进行从小到大的排序

	//for (int i = 1; i <= n; i++) cout << a[i].l << "  " << a[i]. r << endl;

	int sum = 0;  //木板总量
	int mu = -1;  //当前木板扑到的位置

	for (int i = 1; i <= n; i++)
	{
		//情况一：木板铺到的位置包含当前泥坑的终点，用于考虑两个泥坑相邻且第二个泥坑较小的情况
		if (mu >= a[i].r) continue;

		//情况二：木板铺到的位置包含当前泥坑的起点，用于考虑两个泥坑相邻且第二个泥坑较大的情况
		if (mu > a[i].l)
		{
			int len = a[i].r - mu; //需要铺木板的距离
			int shu = len / m;  //得到需要铺的木板量
			if (mu + shu * m < a[i].r) shu++; //若不够则加一
			mu += shu * m; //更新木板铺到的位置
			sum += shu;
		}
		else //情况三：泥坑不相邻
		{
			int len = a[i].r - a[i].l; //需要铺木板的距离
			int shu = len / m;         //得到需要铺的木板量
			if (shu * m < len) shu++;  //若不够则加一
			mu = a[i].l + shu * m;     //更新木板铺到的位置
			sum += shu;
		}
	}
	cout << sum << endl;
	return 0;
}