1588: [HNOI2002]营业额统计

题目链接

题目大意：给出一个n个数的数列a,对于第i个元素ai定义fi=min(abs(ai−aj))，(1<=j<i)，其中f1=a1。输出sum(fi)(1<=i<=n)

题解：稍有常识的人都能看出，这是一道平衡树模板，然而线段树这么神，当然要致辞它啦

我的收获：权值线段树神啊

#include<cstdio>#include<algorithm>#include<map>#define N 50010#define inf 0x7fffffffusing namespace std;int n,a[N],d[N];struct Data{    int x,ord;}data[N];bool cmp(Data a,Data b){    return a.x<b.x;}struct node{    int l,r,sum,mx,mn;}t[N<<2];void pushup(int rt){    t[rt].sum=t[rt<<1].sum+t[rt<<1|1].sum;    t[rt].mx=max(t[rt<<1].mx,t[rt<<1|1].mx);    t[rt].mn=min(t[rt<<1].mn,t[rt<<1|1].mn);}void build(int rt,int l,int r){    t[rt].l=l;t[rt].r=r;t[rt].mx=-1;t[rt].mn=inf-1;    if(l==r) return;    int mid=(l+r)>>1;    if(l<=mid) build(rt<<1,l,mid);    if(r>mid) build(rt<<1|1,mid+1,r);}void modify(int rt,int pos){    if(t[rt].l==t[rt].r){        t[rt].sum=1;        t[rt].mn=t[rt].l;        t[rt].mx=t[rt].l;        return;    }    int mid=(t[rt].l+t[rt].r)>>1;    if(pos<=mid) modify(rt<<1,pos);    if(pos>mid) modify(rt<<1|1,pos);    pushup(rt);}int querymx(int rt,int l,int r){    if(r<l) return -1;    if(t[rt].sum==0) return -1;    if(l<=t[rt].l&&t[rt].r<=r) return t[rt].mx;    int mid=(t[rt].l+t[rt].r)>>1,ret=-2;    if(l<=mid) ret=max(ret,querymx(rt<<1,l,r));    if(r>mid) ret=max(ret,querymx(rt<<1|1,l,r));    return ret;}int querymn(int rt,int l,int r){    if(r<l) return inf-1;    if(t[rt].sum==0) return inf-1;    if(l<=t[rt].l&&t[rt].r<=r) return t[rt].mn;    int mid=(t[rt].l+t[rt].r)>>1,ret=inf;    if(l<=mid) ret=min(ret,querymn(rt<<1,l,r));    if(r>mid) ret=min(ret,querymn(rt<<1|1,l,r));    return ret;}int main(){    scanf("%d",&n);    for(int i=1;i<=n;i++){        if(scanf("%d",&a[i])==EOF) a[i]=0;        data[i].ord=i;data[i].x=a[i];    }    sort(data+1,data+1+n,cmp);    int ord=1;d[1]=data[1].x;a[data[1].ord]=ord;    for(int i=2;i<=n;i++){        if(data[i].x!=data[i-1].x) ord++;        a[data[i].ord]=ord;d[ord]=data[i].x;    }    build(1,1,ord);    int ans=d[a[1]];modify(1,a[1]);    for(int i=2;i<=n;i++){        int x=querymx(1,1,a[i]-1),y=querymn(1,a[i],n),tmp=inf;        if(x!=-1) tmp=min(tmp,d[a[i]]-d[x]);        if(y!=inf-1) tmp=min(tmp,d[y]-d[a[i]]);        ans+=tmp;        modify(1,a[i]);    }    printf("%d\n",ans);    return 0;}

当然，用双向链表维护也是支持的

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>#include <queue>using namespace std;#define INF 0x3f3f3f3fconst int M=100010;int n,rank[M];int nxt[M],pre[M];long long ans;struct data{int id,v;}a[M];bool operator <(data x,data y){return x.v<y.v;}void work(){    for(int i=n;i>1;i--)    {        int x=rank[i],l=INF,r=INF;        if(pre[x]>=1&&pre[x]<=n) l=a[x].v-a[pre[x]].v;        if(nxt[x]>=1&&nxt[x]<=n) r=a[nxt[x]].v-a[x].v;        ans+=min(l,r);        nxt[pre[x]]=nxt[x];        pre[nxt[x]]=pre[x];    }    printf("%lld\n",ans+a[rank[1]].v);}void init(){    scanf("%d",&n);    for(int i=1;i<=n;i++) scanf("%d",&a[i].v),a[i].id=i;    sort(a+1,a+1+n);    for(int i=1;i<=n;i++) rank[a[i].id]=i;    for(int i=1;i<=n;i++) pre[i]=i-1,nxt[i]=i+1; }int main(){    init();    work();    return 0;}