# [BZOJ1689]&&[Usaco2005 Open]Muddy roads 泥泞的路

## 描述 Description

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.

## 输入格式 InputFormat

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.

12 13
15 17
1 3
6 9

1

Silver

BZOJ 1689

## 代码 Code

``````#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int inf=0x7fffffff/11.27;
int i,j,t,n,m,l,r,k,z,y,x,ans;
struct pool
{
int s,e;
bool operator < (const pool &temp) const
{
return (s==temp.s)?e<temp.e:s<temp.s;
}
}a;
inline void read(int &x)
{
x=0;char ch=getchar();
while (ch<'0' || ch>'9') ch=getchar();
while (ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
}
int main()
{
read(n);read(l);
for (i=1;i<=n;i++) read(a[i].s),read(a[i].e);
sort(a+1,a+n+1);
ans=0;t=a.s;k=1;
while (k<=n)
{
while (t+l>=a[k].e) k++;
t=max(t+l,a[k].s);
ans++;
}
printf("%d\n",ans);
return 0;
}``````