NOIP 模拟 $20\; \rm z$

题解

很考验思维的一道题

对于不同的任务点，发现如果 \(x_{i-1}<x_i<x_{i+1}\) 或 \(x_{i-1}>x_i>x_{i+1}\) 那么 \(x_i\) 这个位置的数就没用了

将序列先扫一遍，合并不同的位置，然后将合并后的 \(x_i->x_{i+1}\) 按距离排序，再将询问序列从小到大排序，离线询问

用一个 \(map\) 存储所有 \(x_{i}->x_{i+1}\) 的任务编号，二分查找当前点

那么当一个长度大于这段区间了，它就会超出范围，要将它左右的区间和它合并

注意：ans[c[t].id]=calc(ans[c[t].l)，t++ 不能写成 ans[c[t].id]=calc(ans[c[t++].l)，因为在高版本 c++ 中是从右往左编译的

Code

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++
    template<typename T>inline void read(T &x) {
        ri f=1;x=0;register char ch=gc();
        while(ch<'0'||ch>'9') {if (ch=='-') f=0;ch=gc();}
        while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
        x=f?x:-x;
    }
}
using IO::read;
namespace nanfeng{
    #define node(id,x) (node){id,x}
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    typedef long long ll;
    static const int N=1e5+7;
    struct node{int id;ll x;}al[N];
    inline int operator<(const node &n1,const node &n2) {return n1.x>n2.x;}
    inline int cmp(node n1,node n2) {return n1.x<n2.x;}
    priority_queue<node> que;
    map<int,int> mp;
    int n,q,cnt,t=1;
    ll sum,dx[N],ans[N];
    inline ll calc(ll l) {
        if (mp.empty()) return 0;
        if (mp.begin()->second<0) return sum-(mp.size()-1)*l;
        return sum-mp.size()*l;
    }
    inline int main() {
        // FI=freopen("nanfeng.in","r",stdin);
        // FO=freopen("nanfeng.out","w",stdout);
        read(n),read(q);
        ri lst=0;
        for (ri i(1),x;i<=n;p(i)) {
            read(x);
            if (x==lst) continue;
            if (cnt&&(dx[cnt]<0&&x-lst<0||dx[cnt]>0&&x-lst>0)) dx[cnt]+=x-lst;
            else dx[p(cnt)]=x-lst;
            lst=x;
        }
        for (ri i(1),l;i<=q;p(i)) read(l),al[i].x=l,al[i].id=i;
        sort(al+1,al+q+1,cmp);
        for (ri i(1);i<=cnt;p(i)) {
            sum+=abs(dx[i]);
            mp[i]=dx[i];
            que.push(node(i,abs(dx[i])));
        }
            while(!que.empty()) {
            node tmp=que.top();que.pop();
            auto it=mp.lower_bound(tmp.id);
            if (it==mp.end()) continue;
            node nw=node(it->first,it->second);
            if (abs(nw.x)!=tmp.x||nw.id!=tmp.id) continue;
            while(t<=q&&tmp.x>al[t].x) ans[al[t].id]=calc(al[t++].x);
            auto bg=mp.begin();
            if (it!=mp.begin()) {
                if (it!=prev(mp.end())) {
                    auto pr=prev(it),nx=next(it);
                    node tmpr=node(pr->first,pr->second);
                    node tmpn=node(nx->first,nx->second);
                    mp.erase(pr);mp.erase(nx);
                    sum-=abs(nw.x);
                    sum-=abs(tmpr.x);
                    sum-=abs(tmpn.x);
                    tmp.x=nw.x;
                    tmp.x+=tmpr.x;
                    tmp.x+=tmpn.x;
                    it->second=tmp.x;
                    tmp.x=abs(tmp.x);
                    sum+=tmp.x;
                    que.push(tmp);
                } else {
                    sum-=abs(nw.x);
                    mp.erase(it);
                }
            } else {
                if (nw.x>0) {
                    if (it!=prev(mp.end())) {
                        auto nx=next(it);
                        node tmpn=node(nx->first,nx->second);
                        mp.erase(nx);
                        sum-=abs(nw.x);
                        sum-=abs(tmpn.x);
                        tmp.x=nw.x;
                        tmp.x+=tmpn.x;
                        if (tmp.x) {
                            it->second=tmp.x;
                            tmp.x=abs(tmp.x);
                            sum+=tmp.x;
                            que.push(tmp);
                        } else mp.erase(it);
                    } else {
                        sum-=abs(tmp.x);
                        mp.erase(it);
                    }
                }
            } 
        }
        while(t<=q) ans[al[t].id]=calc(al[t++].x); 
        for (ri i(1);i<=q;p(i)) printf("%lld\n",ans[i]);
        return 0;
    }  
}
int main() {return nanfeng::main();}