题目:http://www.lydsy.com/JudgeOnline/problem.php?id=1150
很明显我们只会连相邻的两幢楼
所以样例就是:2 1 2 6
根据题目意思我们不能取相邻的两个数
用贪心的思想,先取最小的1 然后 删去 2 1 2 再 加上 3 表示不取1而去1的左右两边
但是要注意边界
如 1 2 3 4
很明显如果取1,就一定不会取2,取2不会更优
可以用set来代替堆
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cstdlib> #include<iostream> #include<fstream> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<queue> #include<stack> #include<map> #include<utility> #include<set> #include<bitset> #include<vector> #include<functional> #include<deque> #include<cctype> #include<climits> #include<complex> //#include<bits/stdc++.h>适用于CF,UOJ,但不适用于poj using namespace std; typedef long long LL; typedef double DB; typedef pair<int,int> PII; typedef complex<DB> CP; #define mmst(a,v) memset(a,v,sizeof(a)) #define mmcy(a,b) memcpy(a,b,sizeof(a)) #define re(i,a,b) for(i=a;i<=b;i++) #define red(i,a,b) for(i=a;i>=b;i--) #define fi first #define se second #define m_p(a,b) make_pair(a,b) #define SF scanf #define PF printf #define two(k) (1<<(k)) template<class T>inline T sqr(T x){return x*x;} template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;} template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;} const DB EPS=1e-9; inline int sgn(DB x){if(abs(x)<EPS)return 0;return(x>0)?1:-1;} const DB Pi=acos(-1.0); inline int gint() { int res=0;bool neg=0;char z; for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar()); if(z==EOF)return 0; if(z=='-'){neg=1;z=getchar();} for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar()); return (neg)?-res:res; } inline LL gll() { LL res=0;bool neg=0;char z; for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar()); if(z==EOF)return 0; if(z=='-'){neg=1;z=getchar();} for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar()); return (neg)?-res:res; } const int maxN=100000; int N,K; int dist[maxN+100]; int cnt; struct Tlist { int l,r,id; LL val; inline friend bool operator <(Tlist a,Tlist b){return (a.val!=b.val)?a.val<b.val:a.id<b.id;} } list[2*maxN+100]; set<Tlist> S; LL ans; int main() { freopen("bzoj1150.in","r",stdin); freopen("bzoj1150.out","w",stdout); int i; N=gint();K=gint(); re(i,1,N)dist[i]=gint(); sort(dist+1,dist+N+1); red(i,N,2)dist[i]=dist[i]-dist[i-1]; re(i,1,N-1)dist[i]=dist[i+1]; N--; cnt=N; re(i,1,N) { list[i].id=i; if(i!=1) list[i].l=i-1; if(i!=N) list[i].r=i+1; list[i].val=LL(dist[i]); S.insert(list[i]); } ans=0; while(K--) { int t=S.begin()->id; if(list[t].l==0) { ans+=list[t].val; S.erase(list[t]); S.erase(list[list[t].r]); list[list[list[t].r].r].l=0; } else if(list[t].r==0) { ans+=list[t].val; S.erase(list[t]); S.erase(list[list[t].l]); list[list[list[t].l].l].r=0; } else { ans+=list[t].val; S.erase(list[t]); if(list[t].l!=0)S.erase(list[list[t].l]); if(list[t].r!=0)S.erase(list[list[t].r]); ++cnt; list[cnt].id=cnt; list[cnt].val=list[list[t].l].val+list[list[t].r].val-list[t].val; list[cnt].l=list[list[t].l].l; list[cnt].r=list[list[t].r].r; if(list[cnt].l!=0) list[list[cnt].l].r=cnt; if(list[cnt].r!=0) list[list[cnt].r].l=cnt; S.insert(list[cnt]); } } cout<<ans<<endl; return 0; }