题目链接
题解
真正地理解了一波线段树标记永久化的姿势
每个节点维护两个值(v)和(tag)
(v)代表儿子中的最值
(tag)代表未下传的最值
显然节点的区间大于等于(v)的实际区间
而(tag)的区间包含节点的区间
我们在修改的时候，沿路(v)都要修改，底层(tag)修改
我们在查询的时候，沿路(tag)都要查询，底层(v)查询
(upd:)理解层面上看，和普通线段树是一样的嘛，，，，
#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
",T.query(1,1,D,1,D,1,S));
return 0;
}