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

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.]

题解：简单说一下我的思路

算出相邻放牛的两个牛棚的距离，给它们做降序排序。前M个大的距离两牛棚之间没有模板。简单的贪心。

注：优先队列默认排序是升序，sort函数的排序默认是降序。 所以我两个结构体里的排序函数逻辑不一样。

/*
ID: cxq_xia1
PROG: barn1
LANG: C++
*/
#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
int M,S,C;
struct stall
{
    int num;                                //牛所在牛棚的编号
    bool isIncision;                        //模板在该牛棚 是否分割
    friend bool operator < (stall a,stall b)
    {
        return a.num<b.num;
    }
};

struct node                                  //结构体里存 相邻两个放牛的牛棚里相隔的距离
{
    int IncisionNum;
    int dis;

    friend bool operator < (node a,node b)
    {
        return a.dis<b.dis;
    }
};
int main()
{
    freopen("barn1.in","r",stdin);
    freopen("barn1.out","w",stdout);
    node tmpDis1,tmpDis2;
    stall stall_number[202];
    priority_queue<node> q;
    int ans=0;
    int start;
    bool isStart=true;

    cin >> M >> S >> C;

    cin>>stall_number[0].num;
    stall_number[0].isIncision=false;

    for(int i=1;i<C;i++)
    {
        cin >> stall_number[i].num;
        stall_number[i].isIncision=false;
    }
    sort(stall_number,stall_number+C);

    for(int i=1;i<C;i++)
    {
        tmpDis1.IncisionNum=i-1;
        tmpDis1.dis=stall_number[i].num-stall_number[i-1].num;
        q.push(tmpDis1);
    }
    for(int i=0;i<M-1;i++)
    {
        tmpDis2=q.top();
        q.pop();

        stall_number[tmpDis2.IncisionNum].isIncision=true;
    }

    start=stall_number[0].num;
    for(int i=0;i<C;i++)
    {
        if(!stall_number[i].isIncision)
        {
            if(isStart)
            {
                start=stall_number[i].num;
                isStart=false;
            }
        }
        else
        {
            ans+=(stall_number[i].num-start+1);
            isStart=true;
            start=stall_number[i+1].num;
        }
    }
    ans+=(stall_number[C-1].num-start+1);
    cout << ans << endl;
    return 0;
}