题意： 有n栋楼排成一排，要求在k对楼之间连接电缆，每栋楼只能被连接一次，求最小的∑|dis[x]−dis[y]|.x，y分别为电缆的两个端点。n<=100000
解题方法： 显然k对电缆连接的必然都是相邻的两栋楼，那么问题可以转换成有n-1个数，a[i]=dis[i+1]-dis[i]，从中取出k个数满足其互不相邻且和最小。那么就可以用贪心来做。每次找一个最小的，假设是x，那么x+1和x-1要么不选，要么都选，那么可以把这三个数合并成一个数，权值为a[x+1]+a[x-1]-a[x]，然后把原来的三个数删掉即可。可以用优先队列和双向链表来维护。为了方便的使用priority_queue，直接把a[n]设为-inf就可以了。细节见代码

//bzoj 1150
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x7fffffff;
const int maxn = 100010;
typedef long long LL;
inline int read()
{
    char ch=getchar();
    int f=1,x=0;
    while(!(ch>='0'&&ch<='9')){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}
inline LL llread(){
    char ch=getchar();
    LL f=1,x=0;
    while(!(ch>='0'&&ch<='9')){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}
struct node{
    int pos;
    LL v;
    node(){}
    node(int pos, LL v) : pos(pos), v(v) {}
    bool operator < (const node &rhs) const{
        return v < rhs.v;
    }
};
priority_queue <node> que;
int n, m, L[maxn], R[maxn], vis[maxn];
LL a[maxn], dis[maxn];

int main(){
    n = read(); m = read();
    while(!que.empty()) que.pop();
    for(int i = 1; i <= n; i++) dis[i] = llread();
    for(int i = 1; i <= n; i++){
        if(i < n) a[i] = dis[i] - dis[i + 1];
        else a[i] = -inf;
        node u; u.v = a[i]; u.pos = i;
        que.push(u);
        R[i] = i + 1; L[i] = i - 1;
    }
    R[n] = 1; L[1] = n;
    LL ans = 0;
    for(int i = 1; i <= m; i++){
        node now = que.top(); que.pop();
        while(vis[now.pos] && !que.empty()){now = que.top(); que.pop();}
        ans -= now.v;
        int x = now.pos;
        node v;
        v.pos = x; v.v = a[x] = 1LL * (a[R[x]] + a[L[x]] - a[x]);
        que.push(v);
        R[L[L[x]]] = x;
        L[R[R[x]]] = x;
        vis[L[x]] = 1;
        vis[R[x]] = 1;
        R[x] = R[R[x]];
        L[x] = L[L[x]];
    }
    printf("%lld\n", ans);
    return 0;
}