BZOJ4695:最假女選手
阿新 • • 發佈:2019-01-06
淺談區間最值操作和歷史最值問題:https://www.cnblogs.com/AKMer/p/10225100.html
題目傳送門:https://lydsy.com/JudgeOnline/problem.php?id=4695
吉司機線段樹板子大集合。所有資訊都封裝在一個結構體裡會比開多個數組快14秒。
注意暴力\(dfs\)子樹時要\(pushdown\)。
時間複雜度:\(O(nlog^2n)\)
空間複雜度:\(O(nlogn)\)
程式碼如下:
#include <cstdio> #include <algorithm> using namespace std; typedef long long ll; const int maxn=5e5+5,inf=1e9; int n,m; int a[maxn]; inline int read() { int x=0,f=1;char ch=getchar(); for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1; for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0'; return x*f; } struct segment_tree { struct tree_node { ll sum; int cntA,cntZ,tag,A,B,Y,Z;//A最大值,B嚴格次大值,Y嚴格次小值,Z最小值 }tree[maxn<<2]; inline void update(int p) { tree[p].A=max(tree[p<<1].A,tree[p<<1|1].A); tree[p].Z=min(tree[p<<1].Z,tree[p<<1|1].Z); tree[p].sum=tree[p<<1].sum+tree[p<<1|1].sum; tree[p].cntA=(tree[p<<1].A==tree[p].A)*tree[p<<1].cntA; tree[p].cntA+=(tree[p<<1|1].A==tree[p].A)*tree[p<<1|1].cntA; tree[p].cntZ=(tree[p<<1].Z==tree[p].Z)*tree[p<<1].cntZ; tree[p].cntZ+=(tree[p<<1|1].Z==tree[p].Z)*tree[p<<1|1].cntZ; tree[p].B=tree[p<<1|1].A!=tree[p].A?tree[p<<1|1].A:tree[p<<1|1].B; tree[p].B=max(tree[p].B,tree[p<<1].A!=tree[p].A?tree[p<<1].A:tree[p<<1].B); tree[p].Y=tree[p<<1|1].Z!=tree[p].Z?tree[p<<1|1].Z:tree[p<<1|1].Y; tree[p].Y=min(tree[p].Y,tree[p<<1].Z!=tree[p].Z?tree[p<<1].Z:tree[p<<1].Y); } inline void build(int p,int l,int r) { if(l==r) { tree[p].cntA=tree[p].cntZ=1; tree[p].sum=tree[p].A=tree[p].Z=a[l]; tree[p].B=-inf,tree[p].Y=inf; return; } int mid=(l+r)>>1; build(p<<1,l,mid); build(p<<1|1,mid+1,r); update(p); } inline void add_tag(int p,int l,int r,int v) { tree[p].A+=v,tree[p].Z+=v,tree[p].tag+=v; if(tree[p].Y!=inf)tree[p].Y+=v; if(tree[p].B!=-inf)tree[p].B+=v; tree[p].sum+=1ll*(r-l+1)*v; } inline void max_tag(int p,int v) { tree[p].sum+=1ll*tree[p].cntZ*(v-tree[p].Z),tree[p].Z=v; if(tree[p].Y==inf)tree[p].A=v; else tree[p].B=max(tree[p].B,v); } inline void min_tag(int p,int v) { tree[p].sum-=1ll*tree[p].cntA*(tree[p].A-v),tree[p].A=v; if(tree[p].B==-inf)tree[p].Z=v; else tree[p].Y=min(tree[p].Y,v); } inline void solveMax(int p,int l,int r,int limit) { if(limit<=tree[p].Z)return; if(limit>tree[p].Z&&limit<tree[p].Y) { max_tag(p,limit); return; } int mid=(l+r)>>1;push_down(p,l,r); solveMax(p<<1,l,mid,limit); solveMax(p<<1|1,mid+1,r,limit); update(p); } inline void solveMin(int p,int l,int r,int limit) { if(limit>=tree[p].A)return; if(limit<tree[p].A&&limit>tree[p].B) { min_tag(p,limit); return; } int mid=(l+r)>>1;push_down(p,l,r); solveMin(p<<1,l,mid,limit); solveMin(p<<1|1,mid+1,r,limit); update(p); } inline void push_down(int p,int l,int r) { int mid=(l+r)>>1; if(tree[p].tag) { add_tag(p<<1,l,mid,tree[p].tag); add_tag(p<<1|1,mid+1,r,tree[p].tag); tree[p].tag=0; } solveMin(p<<1,l,mid,tree[p].A); solveMin(p<<1|1,mid+1,r,tree[p].A); solveMax(p<<1,l,mid,tree[p].Z); solveMax(p<<1|1,mid+1,r,tree[p].Z); } inline void ADD(int p,int l,int r,int L,int R,int v) { if(L<=l&&r<=R) { add_tag(p,l,r,v); return; } int mid=(l+r)>>1;push_down(p,l,r); if(L<=mid)ADD(p<<1,l,mid,L,R,v); if(R>mid)ADD(p<<1|1,mid+1,r,L,R,v); update(p); } inline void MAX(int p,int l,int r,int L,int R,int v) { if(tree[p].Z>=v)return; if(L<=l&&r<=R) { solveMax(p,l,r,v); return; } int mid=(l+r)>>1;push_down(p,l,r); if(L<=mid)MAX(p<<1,l,mid,L,R,v); if(R>mid)MAX(p<<1|1,mid+1,r,L,R,v); update(p); } inline void MIN(int p,int l,int r,int L,int R,int v) { if(tree[p].A<=v)return; if(L<=l&&r<=R) { solveMin(p,l,r,v); return; } int mid=(l+r)>>1;push_down(p,l,r); if(L<=mid)MIN(p<<1,l,mid,L,R,v); if(R>mid)MIN(p<<1|1,mid+1,r,L,R,v); update(p); } inline ll querySum(int p,int l,int r,int L,int R) { if(L<=l&&r<=R)return tree[p].sum; int mid=(l+r)>>1;ll res=0;push_down(p,l,r); if(L<=mid)res=querySum(p<<1,l,mid,L,R); if(R>mid)res+=querySum(p<<1|1,mid+1,r,L,R); return res; } inline int queryMax(int p,int l,int r,int L,int R) { if(L<=l&&r<=R)return tree[p].A; int mid=(l+r)>>1,res=-inf;push_down(p,l,r); if(L<=mid)res=max(res,queryMax(p<<1,l,mid,L,R)); if(R>mid)res=max(res,queryMax(p<<1|1,mid+1,r,L,R)); return res; } inline int queryMin(int p,int l,int r,int L,int R) { if(L<=l&&r<=R)return tree[p].Z; int mid=(l+r)>>1,res=inf;push_down(p,l,r); if(L<=mid)res=min(res,queryMin(p<<1,l,mid,L,R)); if(R>mid)res=min(res,queryMin(p<<1|1,mid+1,r,L,R)); return res; } }T; int main() { n=read(); for(int i=1;i<=n;i++) a[i]=read(); T.build(1,1,n); m=read(); for(int i=1;i<=m;i++) { int opt=read(),l=read(),r=read(),v=(opt<4?read():0); if(opt==1)T.ADD(1,1,n,l,r,v); if(opt==2)T.MAX(1,1,n,l,r,v); if(opt==3)T.MIN(1,1,n,l,r,v); if(opt==4)printf("%lld\n",T.querySum(1,1,n,l,r)); if(opt==5)printf("%d\n",T.queryMax(1,1,n,l,r)); if(opt==6)printf("%d\n",T.queryMin(1,1,n,l,r)); } return 0; }