洛谷$P3620\ [APIO/CTSC 2007]$数据备份 贪心

正解:贪心

解题报告:

传送门$QwQ$

$umm$感觉这种问题还蛮经典的,,,就选了某个就不能选另一个这样儿,就可以用堆模拟反悔操作

举个$eg$,如果提出了$a_i$,那就$a_{i-1}$和$a_{i+1}$都不能选了,所以如果选了$a_i$之后想反悔选$a_{i-1}$和$a_{i+1}$就相当于只能另外获得$a_{i+1}+a_{i-1}-a_{i}$的收益了.所以每次取出堆顶$a_{i}$之后,就把$a_{i-1}$和$a_{i+1}$删了,然后插入$a_{i-1}+a_{i+1}-a_{i}$.然后多个数也差不多$QwQ$

所以就用个堆+链表维护下就成$QwQ$

$over$

#include
using namespace std;
#define il inline
#define gc getchar()
#define ll long long
#define ri register int
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)
#define lb(x) lower_bound(st+1,st+st_cnt+1,x)-st

const int N=500000+10,inf=1e9;
int n,K,a[N],l[N],r[N],d[N];
ll as;
bool vis[N];
struct node{ll dat;int id;}tmp;
priority_queueQ;

il int read()
{
    rc ch=gc;ri x=0;rb y=1;
    while(ch!='-' && (ch>'9' || ch<'0'))ch=gc;
    if(ch=='-')ch=gc,y=0;
    while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=gc;
    return y?x:-x;
}
il bool operator < (node gd,node gs){return gd.dat>gs.dat;}

int main()
{
    //freopen("3620.in","r",stdin);freopen("3620.out","w",stdout);
    n=read();K=read();
    rp(i,0,n-1){d[i]=read();if(i){tmp=(node){a[i]=d[i]-d[i-1],i};Q.push(tmp);l[i]=i-1;r[i]=i+1;}}
    r[0]=1;l[n]=n-1;a[0]=a[n]=inf;
    while(K--)
    {
        while(vis[Q.top().id])Q.pop();
        node nw=Q.top();Q.pop();ri pos=nw.id;
        as+=nw.dat;nw.dat=a[pos]=a[l[pos]]+a[r[pos]]-a[pos];Q.push(nw);
        vis[l[pos]]=vis[r[pos]]=1;l[pos]=l[l[pos]];r[pos]=r[r[pos]];r[l[pos]]=pos;l[r[pos]]=pos;
    }
    printf("%lld\n",as);
    return 0;
}

View Code