USACO Barn Repair 解题日志

这道题，一开始没怎么看懂题，在网上看了一些其他人做的才明白了题目的意思。

我们要用尽可能短的木板长度，把有牛的stall给遮起来。

虽然题目说最多提供M块木板，但是想想看就知道，如果用少于M块木板，有些没有牛的、没有必要被遮盖的stall就会被盖住，这样就增长了木板长度。所以，要想得到尽可能短的木板长度，我们使用M块木板。

然后就是考虑怎么遮盖stall的问题了。

因为要使用M块木板，所以会有（M - 1）个大的间隙（大是相对而言的）。

当我们输入有牛的stall号后，最前面的stall和最后面的stall之间的差距就定下来了。（坑爹的是，我开始以为他的输入是从大到小排列的，没想到有数据不是，还需要自己sort一下）所以，要使木板长度最小，间隙们要长度最大。所以只要枚举间隙，找出最大的（M - 1)个，把它们从总差距里减掉就是了。

下面是题目和代码：

Barn Repair

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.

PROGRAM NAME: barn1

INPUT FORMAT

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

SAMPLE INPUT (file barn1.in)

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

OUTPUT FORMAT

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

SAMPLE OUTPUT (file barn1.out)

#include<fstream>
#include<iostream>
#include<algorithm>
using namespace std;

ifstream fin("barn1.in");
ofstream fout("barn1.out");

int main()
{
int M, S, C;

fin >> M >> S >> C;
int stall[200] = {0};
for (int i = 0; i < C; ++i)
fin >> stall[i];

int gap[200] = {0};
int j = 0;
sort(stall, stall + C);
for (int i = 0; i < C - 1; ++i){
gap[j++] = stall[i + 1] - stall[i] - 1;
}

sort(gap, gap + j);

int totalLength = stall[C - 1] - stall[0] + 1;

int k = j - 1;
while (--M){
totalLength -= gap[k--];
}

fout << totalLength << endl;

fin.close();
fout.close();

return 0;
}

25

[One minimum arrangement is one board covering stalls 3-8, one covering 14-21, one covering 25-31, and one covering 40-43.]