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.
从hzw前辈那里搬来的翻译：
牧场里下了一场暴雨，泥泞道路上出现了许多水坑，约翰想用一批长度为L的木板将这些水坑盖住. 牧场里的道路可以看成一根数轴，每个水坑可以用数轴上的两个坐标表示，如(3，6)表示从3到6有一个长度为3的水坑．所有的水坑都是不重叠的，(3，6)和(6，9)可以出现在同一个输入数据中，因为它们是两个连续的水坑，但不重叠．
请你帮助约翰计算最少要用多少块木板才能将所有水坑盖住
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 Input
3 3 1 6 13 17 8 12
Sample Output
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
Source
贪心水题。输入的范围排好序以后挨个铺木板就行。
1 #include<iostream>
2 #include<algorithm>
3 #include<cstring>
4 #include<cstdio>
5 #include<cmath>
6 using namespace std;
7 const int mxn=12000;
8 struct wd{
9 int x,y;
10 bool operator < (const wd rhd){
11 return x<rhd.x;
12 }
13 }a[mxn];
14 int n,L;
15 int main(){
16 scanf("%d%d",&n,&L);
17 int i,j;
18 for(i=1;i<=n;i++){
19 scanf("%d%d",&a[i].x,&a[i].y);
20 }
21 sort(a+1,a+n+1);
22 int cnt=0;
23 int now=0;
24 for(i=1;i<=n;i++){
25 now=max(now,a[i].x);//开始铺
26 while(now<a[i].y){
27 now+=L;
28 cnt++;
29 }
30 }
31 printf("%d\n",cnt);
32 return 0;
33 }