USACO 1.2 Barn Repair 贪心 牛棚问题

题目连接

http://train.usaco.org/usacoprob2?a=fnrIvV5X91K&S=barn1

Description

It was a dark and stormy night that ripped the roof and gates off the stalls that hold Farmer John’s cows. Happily, many of the cows were on vacation, so the barn was not completely full.

The cows spend the night in stalls that are arranged adjacent to each other in a long line. Some stalls have cows in them; some do not. All stalls are the same width.

Farmer John must quickly erect new boards in front of the stalls, since the doors were lost. His new lumber supplier will supply him boards of any length he wishes, but the supplier can only deliver a small number of total boards. Farmer John wishes to minimize the total length of the boards he must purchase.

Given M (1 <= M <= 50), the maximum number of boards that can be purchased; S (1 <= S <= 200), the total number of stalls; C (1 <= C <= S) the number of cows in the stalls, and the C occupied stall numbers (1 <= stall_number <= S), calculate the minimum number of stalls that must be blocked in order to block all the stalls that have cows in them.

Print your answer as the total number of stalls blocked.

Input

Line 1: M, S, and C (space separated)
Lines 2-C+1: Each line contains one integer, the number of an occupied stall.

Output

A single line with one integer that represents the total number of stalls blocked.

Sample Input

4 50 18
3
4
6
8
14
15
16
17
21
25
26
27
30
31
40
41
42
43

Sample Output

25

题意

可以任意使用任何规格的木板,给出牛棚的位置,用最少长度的m块模板去覆盖所有牛棚。

题解

贪心之前练得真的不多,本身也不喜欢做贪心,所以水平可能因此提不上去吧。这题的思路比较简单,算一道经典题。先用一块大木板去覆盖整个最左边和最右边的牛棚,容易知道使用所有m块木板是最优策略,因为如果少使用一块的话必定多一块木板覆盖不用覆盖的区间。然后m块木板必定产生m-1个空隙,这个就是我们可以优化掉的空间,每次选取最大的缝隙即可,直到选取到m-1个。最后答案就是总长度减去最左边不用覆盖的区间和最右边不用覆盖的区间,再减去刚才选取的优化区间。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
/*
ID: hyson601
PROG: barn1
LANG: C++11
*/
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 210;
int m,s,c,board[maxn],pos[maxn];
bool cmp(int a,int b){
return a>b;
}
int main(){
freopen("barn1.in","r",stdin);
freopen("barn1.out","w",stdout); scanf("%d%d%d",&m,&s,&c);
for(int i=0;i<c;++i) scanf("%d",&pos[i]);
sort(pos,pos+c);
for(int i=1;i<c;++i) board[i]=pos[i]-pos[i-1]-1;
sort(board+1,board+c,cmp);
int cnt=1,ans=0;
for(int i=1;i<c && cnt!=m;++i){
ans+=board[i];
++cnt;
}
printf("%d\n",s-ans-(pos[0]-1)-(s-pos[c-1]));
}