USACO-section1.3 Barn Repair
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题意:
John 需要为牛棚安装木板来看住牛,木板长度不限但是数量有限。
给出可能买到的木板最大的数目M(1<= M<=50);
牛棚的总数S(1<= S<=200);
牛棚里牛的总数C(1 <= C <=S);
和牛所在的牛棚的编号stall_number(1 <= stall_number <= S),计算拦住所有有牛的牛棚所需木板的最小总长度 输出所需木板的最小总长度作为答案
题解:
1前提:要使总长度小,就要使用尽量多的木板。
2所有牛按编号排序(虽然样例中是排好序的);
假设从第一头牛到最后一头牛用一块木板;
相邻两头牛之间间距排序,找出前M-1大的间距和去除,就能得到最小总长度;
3注意:如果M大于C,一头牛一块木板,直接输出C,
/*ID:jsntrdy1PROG: barn1LANG: C++*/#include<cstdio>#include<iostream>#include<cstring>#include<fstream>#include<algorithm>const int N=210;using namespace std;ifstream fin("barn1.in");ofstream fout("barn1.out");int main(){ int m,s,c,res; int a[N];//c头牛所在牛棚号 int b[N];//记录相邻两头牛的距离 fin>>m>>s>>c; if(m>c) { fout<<c<<endl; return 0; } for(int i=0;i<c;i++) { fin>>a[i]; } sort(a,a+c);//*************** for(int i=0;i<c-1;i++) { b[i]=a[i+1]-a[i]; } sort(b,b+c-1); res=a[c-1]-a[0]+m; for(int i=c-2;i>=c-m;i--) { res-=b[i]; } fout<<res<<endl; return 0;}
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