USACO-Section 1.3 Barn Repair（贪心）
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)
25
[One minimum arrangement is one board covering stalls 3-8, one covering 14-21, one covering 25-31, and one covering 40-43.]
做法在Section 1.3第一个TEXT已经说明，就是简单的贪心，每次将空隙最小的两个牛棚连起来用一块木板即可，最后只要用的木板数<=m即可。
注意输入的数据不是按顺序的，需要排序
/*
ID: your_id_here
PROG: barn1
LANG: C++
*/
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
int main() {
int i,tmp,ans,m,s,c,num[205];
priority_queue<int> q;
freopen("barn1.in","r",stdin);
freopen("barn1.out","w",stdout);
while(3==scanf("%d%d%d",&m,&s,&c)) {
ans=c;//刚开始所需的木板数等于c
for(i=0;i<ans;++i)
scanf("%d",num+i);
sort(num,num+ans);
for(i=1;i<ans;++i)
if((tmp=num[i]-num[i-1])!=1)
q.push(1-tmp);//如果两个牛棚不相邻，则将它们之间的空隙的相反数压入队列
else
--c;//如果两个牛棚相邻，则可以用一块木板，即所需木板-1
while(c>m) {//如果所需木板>m，则选空隙最小的连起来，则所需木板-1
ans-=q.top();
q.pop();
--c;
}
printf("%d\n",ans);
}
return 0;
}