Muddy roads（贪心）

Muddy roads

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3288 Accepted: 1511

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.

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 31 613 178 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

USACO 2005 U S Open Silver

题意：

有n摊泥，长度为L，求最少需要多少目板才能覆盖这些泥？

思路：

贪心思想，对每个板的起点由小到大进行排序，然后就可以得出一个最小的值。注意终点 a[i].e-1;

代码：

# include <iostream>#include <stdio.h>#include <cstring>#include <iomanip>#include <cmath>#include <algorithm> #define exp 1e-6typedef long long ll;using namespace std; struct node{int s;int e;}a[10010];bool cmp(node a,node b){return a.s<b.s;}int main(){int n,L,i,j;ios::sync_with_stdio(false);cin>>n>>L;for(i=0;i<n;i++)cin>>a[i].s>>a[i].e;sort(a,a+n,cmp);int sum=0;int last=-1;for(i=0;i<n;i++){if(last>=a[i].e)continue;if(last>a[i].s){int length=a[i].e-last;int num=(length+L-1)/L;last+=num*L;sum+=num;}else{int length=a[i].e-a[i].s;int num=(length+L-1)/L;last=a[i].s+num*L;sum+=num;}}cout<<sum<<endl;    return 0;}