BZOJ.3064.CPU监控(线段树 历史最值)

题目链接
\(Description\)
有一个长为n的序列Ai，要求支持查询[l,r]的最值、历史最值，区间加/重设
\(Solution\)
线段树，每个点再维护一个历史(从0到现在)最大值、历史(从上次下传标记到现在)最大的set,add标记
PushDown时肯定是先下放历史标记，之后再用当前标记更新

/*
要记得当要PushDown某个点时，last,now的val都是历史的(下传前),所以now.v + last.add就是下传前值+[下传前到现在]一次最大的修改的值
不能只在Set后清空add。像代码这么做很正确，可以稍微画个图
每次now的操作不仅要更新now，还要更新last
*/
#include<cstdio>
#include<cctype>
#include<algorithm>
#define gc() getchar()
#define lson node[node[rt].ls]
#define rson node[node[rt].rs]
const int N=1e5+5,INF=0x3f3f3f3f;

int n,m;
inline int read()
{
    int now=0,f=1;register char c=gc();
    for(;!isdigit(c);c=gc()) if(c=='-') f=-1;
    for(;isdigit(c);now=now*10+c-'0',c=gc());
    return now*f;
}
#define now node[rt]
struct Seg_Tree
{
    int tot;
    struct Node
    {
        int ls,rs,nv,lv,nadd,ladd,nset,lset;
    }node[N<<1];
    inline void PushUp(int rt)
    {
        now.nv = std::max(lson.nv, rson.nv);
        now.lv = std::max(lson.lv, rson.lv);
    }
    void Build(int l,int r)
    {
        int rt=tot++;
        now.nadd=now.ladd=0, now.nset=now.lset=-INF;
        if(l==r) now.nv=now.lv=read();
        else
        {
            int m=l+r>>1;
            now.ls=tot, Build(l,m);
            now.rs=tot, Build(m+1,r);
            PushUp(rt);
        }
    }
    inline void Update(int &x,int y) {x = std::max(x,y);}
    inline void L_Set(int rt,int v)
    {
        Update(now.lv, v);
        Update(now.lset, v);
    }
    inline void L_Add(int rt,int v)
    {
        Update(now.lv, now.nv+v);
        if(now.nset>-INF) Update(now.lset, now.nset+v);//判断的是nset>-INF(用now更新last关last什么事)//necessary!
        else Update(now.ladd, now.nadd+v);
        //如果之后rt.ladd下传了，那就没了；如果没下传，仍是rt下传前情况(历史)，即可以用now.ladd;或是用上次下传前标记+到现在修改最大值
    }
    inline void N_Set(int rt,int v)
    {
        Update(now.lv, now.nv=v);
        Update(now.lset, now.nset=v);
        now.nadd = 0;
    }
    inline void N_Add(int rt,int v)
    {
        Update(now.lv, now.nv+=v);
        if(now.nset>-INF) Update(now.lset, now.nset+=v);
        //Update的是lset,not ladd,现在更新的是set不是add//在被set影响后，所有操作都可看做set
        else Update(now.ladd, now.nadd+=v);
    }
    void PushDown(int rt)
    {
        if(now.ladd)
            L_Add(now.ls,now.ladd),L_Add(now.rs,now.ladd),now.ladd=0;
        if(now.lset>-INF)
            L_Set(now.ls,now.lset),L_Set(now.rs,now.lset),now.lset=-INF;
        if(now.nadd)
            N_Add(now.ls,now.nadd),N_Add(now.rs,now.nadd),now.nadd=0;
        if(now.nset>-INF)
            N_Set(now.ls,now.nset),N_Set(now.rs,now.nset),now.nset=-INF;
    }
    void Modify_Add(int l,int r,int rt,int L,int R,int v)
    {
        if(L<=l && r<=R) {N_Add(rt,v); return;}
        PushDown(rt);
        int m=l+r>>1;
        if(L<=m) Modify_Add(l,m,now.ls,L,R,v);
        if(m<R) Modify_Add(m+1,r,now.rs,L,R,v);
        PushUp(rt);
    }
    void Modify_Set(int l,int r,int rt,int L,int R,int v)
    {
        if(L<=l && r<=R) {N_Set(rt,v); return;}
        PushDown(rt);
        int m=l+r>>1;
        if(L<=m) Modify_Set(l,m,now.ls,L,R,v);
        if(m<R) Modify_Set(m+1,r,now.rs,L,R,v);
        PushUp(rt);
    }
    int Query(int l,int r,int rt,int L,int R,bool opt)
    {
        if(L<=l && r<=R) return opt?now.nv:now.lv;
        PushDown(rt);
        int m=l+r>>1;
        if(L<=m)
            if(m<R) return std::max(Query(l,m,now.ls,L,R,opt),Query(m+1,r,now.rs,L,R,opt));
            else return Query(l,m,now.ls,L,R,opt);
        return Query(m+1,r,now.rs,L,R,opt);
    }
}t;
#undef now

int main()
{
    n=read(), t.Build(1,n), m=read();
    char opt[3]; int x,y,z;
    while(m--)
    {
        scanf("%s",opt), x=read(),y=read();
        switch(opt[0])
        {
            case 'Q': printf("%d\n",t.Query(1,n,0,x,y,1)); break;
            case 'A': printf("%d\n",t.Query(1,n,0,x,y,0)); break;
            case 'P': z=read(), t.Modify_Add(1,n,0,x,y,z); break;
            case 'C': z=read(), t.Modify_Set(1,n,0,x,y,z); break;
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/SovietPower/p/8442661.html