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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#262403 | #5069. Vacation | WinResearcher | TL | 0ms | 0kb | C++14 | 5.2kb | 2023-11-23 19:16:49 | 2023-11-23 19:16:50 |
answer
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define REPG(i,h,x) for(int i=h[x];~i;i=edge[i].next)
#define CLR(a,x) memset(a,x,sizeof(a))
#define pii pair<int,int>
const int INF=0x3f3f3f3f;
const ll LLINF=0x3f3f3f3f3f3f3f3f;
const int N=1e6+5;
int n,q,K,bcnt,a[N<<1],st[N],ed[N],bel[N<<1];
ll c[N],d[N],pre[N<<1],suf[N<<1];
struct node{ll pre,suf,sum,mx;};
struct node2{ll v1,v2,mx;};
inline int lc(int p){return p<<1;}
inline int rc(int p){return p<<1|1;}
struct SGT1{
vector<node> t;
inline void init(int n){t.resize(n*4);}
inline node merge(node l,node r){
node ret;
ret.sum=l.sum+r.sum;
ret.pre=max(l.pre,l.sum+r.pre);
ret.suf=max(r.suf,l.suf+r.sum);
ret.mx=max(max(l.mx,r.mx),l.suf+r.pre);
return ret;
}
inline void pushup(int p){t[p]=merge(t[lc(p)],t[rc(p)]);}
void update(int tt,int l,int r,int p,int c){
if(l==r) return (void)(t[p]={max(c,0),max(c,0),c,max(c,0)});
int mid=l+r>>1;
if(tt<=mid) update(tt,l,mid,lc(p),c);
else update(tt,mid+1,r,rc(p),c);
pushup(p);
}
node query(int tl,int tr,int l,int r,int p){
if(tl<=l&&r<=tr) return t[p];
int mid=l+r>>1;
if(tl>mid) return query(tl,tr,mid+1,r,rc(p));
else if(tr<=mid) return query(tl,tr,l,mid,lc(p));
return merge(query(tl,tr,l,mid,lc(p)),query(tl,tr,mid+1,r,rc(p)));
}
void build(int p,int l,int r,int a[]){
if(l==r) return (void)(t[p]={max(a[l],0),max(a[l],0),a[l],max(a[l],0)});
int mid=l+r>>1;
build(lc(p),l,mid,a),build(rc(p),mid+1,r,a);
pushup(p);
}
}t1[N],t5;
struct SGT2{
vector<node2> t;
vector<ll> lazy1,lazy2;
inline void init(int n){t.resize(n*4),lazy1.resize(n*4),lazy2.resize(n*4);}
inline node2 merge(node2 l,node2 r){
node2 ret;
ret.v1=max(l.v1,r.v1),ret.v2=max(l.v2,r.v2);
ret.mx=max(max(l.mx,r.mx),l.v1+r.v2);
return ret;
}
inline void pushup(int p){t[p]=merge(t[lc(p)],t[rc(p)]);}
inline void push(int p,ll v1,ll v2){t[p].v1+=v1,t[p].v2+=v2,t[p].mx+=v1+v2,lazy1[p]+=v1,lazy2[p]+=v2;}
inline void pushdown(int p){if(lazy1[p]||lazy2[p]) push(lc(p),lazy1[p],lazy2[p]),push(rc(p),lazy1[p],lazy2[p]),lazy1[p]=lazy2[p]=0;}
void update(int tl,int tr,int l,int r,int p,ll v1,ll v2){
if(tl<=l&&r<=tr) return push(p,v1,v2);
int mid=l+r>>1;
pushdown(p);
if(tl<=mid) update(tl,tr,l,mid,lc(p),v1,v2);
if(tr>mid) update(tl,tr,mid+1,r,rc(p),v1,v2);
pushup(p);
}
node2 query(int tl,int tr,int l,int r,int p){
if(tl<=l&&r<=tr) return t[p];
int mid=l+r>>1;
if(tl>mid) return query(tl,tr,mid+1,r,rc(p));
else if(tr<=mid) return query(tl,tr,l,mid,lc(p));
return merge(query(tl,tr,l,mid,lc(p)),query(tl,tr,mid+1,r,rc(p)));
}
void build(int p,int l,int r,ll a[],ll b[]){
if(l==r) return (void)(t[p]={a[l],b[l],-LLINF});//v1:pre v2:suf
int mid=l+r>>1;
build(lc(p),l,mid,a,b),build(rc(p),mid+1,r,a,b);
pushup(p);
}
}t2[N];
struct SGT3{
ll t[N<<2];
inline void pushup(int p){t[p]=max(t[lc(p)],t[rc(p)]);}
void update(int tt,int l,int r,int p,ll c){
if(l==r) return (void)(t[p]=c);
int mid=l+r>>1;
if(tt<=mid) update(tt,l,mid,lc(p),c);
else update(tt,mid+1,r,rc(p),c);
pushup(p);
}
ll query(int tl,int tr,int l,int r,int p){
if(tl<=l&&r<=tr) return t[p];
int mid=l+r>>1;ll ret=-LLINF;
if(tl<=mid) ret=query(tl,tr,l,mid,lc(p));
if(tr>mid) ret=max(ret,query(tl,tr,mid+1,r,rc(p)));
return ret;
}
void build(int p,int l,int r,ll a[]){
if(l==r) return (void)(t[p]=a[l]);
int mid=l+r>>1;
build(lc(p),l,mid,a),build(rc(p),mid+1,r,a);
pushup(p);
}
}t3,t4;
int main(){
// ios::sync_with_stdio(false);
// cin.tie(0);
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
scanf("%d%d%d",&n,&q,&K);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
t5.init(n),t5.build(1,1,n,a);
bcnt=(n+K-1)/K+1;
for(int i=1;i<=bcnt;i++){
st[i]=(i-1)*K+1,ed[i]=(i==bcnt?n+K:i*K);
for(int j=st[i];j<=ed[i];j++) bel[j]=i;
printf("%d %d\n",st[i],ed[i]);
t1[i].init(K),t1[i].build(1,1,K,a+st[i]-1);
c[i]=t1[i].t[1].mx;
if(i==1) continue;
for(int j=1;j<=K;j++) pre[j]=pre[j-1]+a[j+st[i]-1];
for(int j=K;j>=1;j--) suf[j]=suf[j+1]+a[j+st[i]-1-K];
t2[i].init(K),t2[i].build(1,1,K,pre,suf);
d[i]=t2[i].t[1].mx;
}
t3.build(1,1,bcnt,c),t4.build(1,1,bcnt,d);
while(q--){
int op,x,y;
scanf("%d%d%d",&op,&x,&y);
if(op==1){
t5.update(x,1,n,1,y);
t1[bel[x]].update(x-st[bel[x]]+1,1,K,1,y),t3.update(bel[x],1,bcnt,1,t1[bel[x]].t[1].mx);
if(bel[x]>1) t2[bel[x]].update(x-st[bel[x]]+1,K,1,K,1,y-a[x],0),t4.update(bel[x],1,bcnt,1,t2[bel[x]].t[1].mx);
if(bel[x]<bcnt) t2[bel[x]+1].update(x-st[bel[x]]+1,K,1,K,1,0,y-a[x]),t4.update(bel[x]+1,1,bcnt,1,t2[bel[x]+1].t[1].mx);
a[x]=y;
}
if(op==2){
if(bel[x]==bel[y]){printf("%lld\n",t5.query(x,y,1,n,1).mx);continue;}
ll ans=0;
ans=max(ans,max(t5.query(x,x+K-1,1,n,1).mx,t5.query(y-K+1,y,1,n,1).mx));
if(bel[x]+1==bel[y]) ans=max(ans,t2[bel[y]].query(x-st[bel[x]]+1,y-st[bel[y]]+1,1,K,1).mx);
else{
ans=max(ans,t3.query(bel[x]+1,bel[y]-1,1,bcnt,1));
ans=max(ans,max(t2[bel[x]+1].query(x-st[bel[x]]+1,K,1,K,1).mx,t2[bel[y]].query(1,y-st[bel[y]]+1,1,K,1).mx));
if(bel[x]+2<=bel[y]) ans=max(ans,t4.query(bel[x]+2,bel[y]-1,1,bcnt,1));
}
printf("%lld\n",ans);
}
}
return 0;
}
詳細信息
Test #1:
score: 0
Time Limit Exceeded
input:
5 6 3 0 -5 -3 8 -3 2 3 5 1 2 5 2 1 5 1 4 -3 2 3 5 2 1 5