題目鏈接

BZOJ1513

題解

真正地理解了一波線段樹標記永久化的姿勢
每個節點維護兩個值\(v\)和\(tag\)
\(v\)代表兒子中的最值
\(tag\)代表未下傳的最值

顯然節點的區間大於等於\(v\)的實際區間
而\(tag\)的區間包含節點的區間

我們在修改的時候，沿路\(v\)都要修改，底層\(tag\)修改

我們在查詢的時候，沿路\(tag\)都要查詢，底層\(v\)查詢

\(upd:\)理解層面上看，和普通線段樹是一樣的嘛，，，，

#include<algorithm>
#include<iostream>
#include<cstring>
 

#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
 

#define ls (u << 1)
#define rs (u << 1 | 1)
using namespace std;
const int maxn = 4005,maxm = 5000005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57
 
){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int n,D,S;
struct segx{
    int mx[maxm],lc[maxm],rc[maxm],tag[maxm],cnt;
    void modify(int& u,int l,int r,int L,int R,int v){
        if (!u) u = ++cnt; mx[u] = max(mx[u],v);
        if (l >= L && r <= R){tag[u] = max(tag[u],v); return;}
        int mid = l + r >> 1;
        if (mid >= L) modify(lc[u],l,mid,L,R,v);
        if (mid < R) modify(rc[u],mid + 1,r,L,R,v);
    }
    int query(int u,int l,int r,int L,int R){
        if (!u) return 0;
        if (l >= L && r <= R) return mx[u];
        int mid = l + r >> 1,ans = tag[u];
        if (mid >= R) return max(ans,query(lc[u],l,mid,L,R));
        if (mid < L) return max(ans,query(rc[u],mid + 1,r,L,R));
        return max(ans,max(query(lc[u],l,mid,L,R),query(rc[u],mid + 1,r,L,R)));
    }
}Seg;
struct segy{
    int rt[maxn],tag[maxn];
    void modify(int u,int l,int r,int L,int R,int ll,int rr,int v){
        Seg.modify(rt[u],1,S,ll,rr,v);
        if (l >= L && r <= R){Seg.modify(tag[u],1,S,ll,rr,v); return;}
        int mid = l + r >> 1;
        if (mid >= L) modify(ls,l,mid,L,R,ll,rr,v);
        if (mid < R) modify(rs,mid + 1,r,L,R,ll,rr,v);
    }
    int query(int u,int l,int r,int L,int R,int ll,int rr){
        if (l >= L && r <= R) return Seg.query(rt[u],1,S,ll,rr);
        int mid = l + r >> 1,ans = Seg.query(tag[u],1,S,ll,rr);
        if (mid >= L) ans = max(ans,query(ls,l,mid,L,R,ll,rr));
        if (mid < R) ans = max(ans,query(rs,mid + 1,r,L,R,ll,rr));
        return ans;
    }
}T;
int main(){
    D = read(); S = read(); n = read();
    int d,s,h,x0,y0,x,y,xx,yy;
    while (n--){
        d = read(); s = read(); h = read(); x0 = read(); y0 = read();
        x = x0 + 1; y = y0 + 1; xx = x0 + d; yy = y0 + s;
        h += T.query(1,1,D,x,xx,y,yy);
        T.modify(1,1,D,x,xx,y,yy,h);
    }
    printf("%d\n",T.query(1,1,D,1,D,1,S));
    return 0;
}

BZOJ1513 [POI2006]Tet-Tetris 3D 【二維線段樹】