題目鏈接
\(Description\)
有一個長為n的序列Ai，要求支持查詢[l,r]的最值、歷史最值，區間加/重設
\(Solution\)
線段樹，每個點再維護一個歷史(從0到現在)最大值、歷史(從上次下傳標記到現在)最大的set,add標記
PushDown時肯定是先下放歷史標記，之後再用當前標記更新

/*
要記得當要PushDown某個點時，last,now的val都是歷史的(下傳前),所以now.v + last.add就是下傳前值+[下傳前到現在]一次最大的修改的值 
不能只在Set後清空add。像代碼這麽做很正確，可以稍微畫個圖 
每次now的操作不僅要更新now，還要更新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);//判斷的是nset>-INF(用now更新last關last什麽事)//necessary!
        else Update(now.ladd, now.nadd+v);
        //如果之後rt.ladd下傳了，那就沒了；如果沒下傳，仍是rt下傳前情況(歷史)，即可以用now.ladd;或是用上次下傳前標記+到現在修改最大值 
    }
    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,現在更新的是set不是add//在被set影響後，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;
}

BZOJ.3064.CPU監控(線段樹 歷史最值)