[可持久化线段树][主席树]小结
前言
今天机房莫名掀起了一股学习主席树的浪潮，然而这对于早几天接触主席树的我，岂不是个XXX的好机会= =
然而我是一个正直的人，所以就把这些鬼东西扔出来，顺便给自己做个小结
静态主席树
顾名思义，就是静态的，不带修改的主席树
很简单的东西，用前缀和维护的，可以查询区间$k$小值的数据结构，然而只有理解了，才可以灵活运用，乃至将其作用推广
懒得写教程了（其实只是不想码字= =）（那不还是懒吗）
当时太年轻，打的太丑= =
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 using namespace std; 6 //#define mem(x) memset((x),0,sizeof(x)) 7 inline int read(){ 8 int sum(0); 9 char ch(getchar()); 10 for(;ch<'0'||ch>'9';ch=getchar()); 11 for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 12 return sum; 13 } 14 int T; 15 int n,m; 16 int a[100005],num[100005]; 17 int cnt,size; 18 int rt[2000005],sum[2000005],lch[2000005],rch[2000005]; 19 inline void clear(){ 20 // mem(a),mem(num),mem(rt),mem(sum),mem(lch),mem(rch); 21 cnt=size=n=m=0; 22 } 23 inline void build(int &x,int l,int r){ 24 x=++cnt; 25 sum[x]=0; 26 if(l==r){ 27 lch[x]=rch[x]=0; 28 return; 29 } 30 int mid((l+r)>>1); 31 build(lch[x],l,mid); 32 build(rch[x],mid+1,r); 33 } 34 inline void update(int &x,int las,int pos,int l,int r){ 35 x=++cnt; 36 lch[x]=lch[las]; 37 rch[x]=rch[las]; 38 sum[x]=sum[las]+1; 39 if(l==r) 40 return; 41 int mid((l+r)>>1); 42 if(pos<=mid) 43 update(lch[x],lch[las],pos,l,mid); 44 else 45 update(rch[x],rch[las],pos,mid+1,r); 46 } 47 inline int query(int ll,int rr,int l,int r,int k){ 48 if(l==r) 49 return l; 50 int mid((l+r)>>1); 51 int cnt(sum[lch[rr]]-sum[lch[ll]]); 52 if(k<=cnt) 53 return query(lch[ll],lch[rr],l,mid,k); 54 return query(rch[ll],rch[rr],mid+1,r,k-cnt); 55 } 56 int main(){ 57 T=read(); 58 while(T--){ 59 clear(); 60 n=read(),m=read(); 61 for(int i=1;i<=n;++i) 62 num[i]=a[i]=read(); 63 sort(num+1,num+n+1); 64 size=unique(num+1,num+n+1)-num-1; 65 for(int i=1;i<=n;++i) 66 a[i]=lower_bound(num+1,num+size+1,a[i])-num; 67 build(rt[0],1,size); 68 for(int i=1;i<=n;++i) 69 update(rt[i],rt[i-1],a[i],1,size); 70 for(int i=1;i<=m;++i){ 71 int l(read()),r(read()),k(read()); 72 int ans(query(rt[l-1],rt[r],1,size,k)); 73 printf("%d\n",num[ans]); 74 } 75 } 76 }
树上建主席树，二分查即可
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 using namespace std; 6 inline int read(){ 7 int sum(0); 8 char ch(getchar()); 9 for(;ch<'0'||ch>'9';ch=getchar()); 10 for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 11 return sum; 12 } 13 struct edge{ 14 int e; 15 edge *n; 16 }a[200005],*pre[100005]; 17 int tot; 18 inline void insert(int s,int e){ 19 a[++tot].e=e; 20 a[tot].n=pre[s]; 21 pre[s]=&a[tot]; 22 } 23 int timee; 24 int dep[100005],fa[100005][20],l[100005],r[100005],pos[100005]; 25 int n,m; 26 int cnt,size; 27 int v[100005],num[100005]; 28 int rt[100005],lch[2000005],rch[2000005],sum[2000005]; 29 inline void dfs(int u){ 30 l[u]=++timee; 31 pos[timee]=u; 32 for(int i=1;(1<<i)<=dep[u];++i) 33 fa[u][i]=fa[fa[u][i-1]][i-1]; 34 for(edge *i=pre[u];i;i=i->n){ 35 int e(i->e); 36 if(e!=fa[u][0]){ 37 fa[e][0]=u; 38 dep[e]=dep[u]+1; 39 dfs(e); 40 } 41 } 42 r[u]=timee; 43 } 44 inline int lca(int x,int y){ 45 if(dep[x]<dep[y]) 46 swap(x,y); 47 int delta(dep[x]-dep[y]); 48 for(int i=0;delta>0;++i) 49 if(delta&(1<<i)){ 50 delta^=1<<i; 51 x=fa[x][i]; 52 } 53 if(x==y) 54 return x; 55 for(int i=19;i>=0;--i) 56 if(fa[x][i]!=fa[y][i]) 57 x=fa[x][i],y=fa[y][i]; 58 return fa[x][0]; 59 } 60 inline void build(int &x,int l,int r){ 61 x=++cnt; 62 sum[x]=0; 63 if(l==r) 64 return; 65 int mid((l+r)>>1); 66 build(lch[x],l,mid); 67 build(rch[x],mid+1,r); 68 } 69 inline void update(int &x,int las,int pos,int l,int r){ 70 x=++cnt; 71 lch[x]=lch[las]; 72 rch[x]=rch[las]; 73 sum[x]=sum[las]+1; 74 if(l==r) 75 return; 76 int mid((l+r)>>1); 77 if(pos<=mid) 78 update(lch[x],lch[las],pos,l,mid); 79 else 80 update(rch[x],rch[las],pos,mid+1,r); 81 } 82 inline int query(int x,int y,int k){ 83 int a(x),b(y),c(lca(x,y)),d(fa[c][0]); 84 a=rt[l[a]],b=rt[l[b]],c=rt[l[c]],d=rt[l[d]]; 85 int l(1),r(size); 86 while(l<r){ 87 int mid((l+r)>>1); 88 int tmp(sum[lch[a]]+sum[lch[b]]-sum[lch[c]]-sum[lch[d]]); 89 if(k<=tmp) 90 r=mid,a=lch[a],b=lch[b],c=lch[c],d=lch[d]; 91 else 92 k-=tmp,l=mid+1,a=rch[a],b=rch[b],c=rch[c],d=rch[d]; 93 } 94 return num[l]; 95 } 96 int ans; 97 int main(){ 98 memset(pre,NULL,sizeof(pre)); 99 n=read(),m=read(); 100 for(int i=1;i<=n;++i) 101 num[i]=v[i]=read(); 102 for(int i=1;i<n;++i){ 103 int x(read()),y(read()); 104 insert(x,y),insert(y,x); 105 } 106 sort(num+1,num+n+1); 107 size=unique(num+1,num+n+1)-num-1; 108 for(int i=1;i<=n;++i) 109 v[i]=lower_bound(num+1,num+size+1,v[i])-num; 110 dfs(1); 111 build(rt[0],1,size); 112 for(int i=1;i<=n;++i) 113 update(rt[i],rt[l[fa[pos[i]][0]]],v[pos[i]],1,size); 114 while(m--){ 115 int x(read()),y(read()),k(read()); 116 x^=ans; 117 ans=query(x,y,k); 118 if(!m) 119 printf("%d",ans); 120 else 121 printf("%d\n",ans); 122 } 123 }
[Poi2014]Couriers
太水，没写博客，基本上就是板子
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 inline int read(){ 6 int sum(0); 7 char ch(getchar()); 8 for(;ch<'0'||ch>'9';ch=getchar()); 9 for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 10 return sum; 11 } 12 int n,m; 13 int cnt; 14 int a[500005]; 15 int rt[15000005],sum[15000005],lch[15000005],rch[15000005]; 16 inline void build(int &x,int l,int r){ 17 x=++cnt; 18 sum[x]=0; 19 if(l==r) 20 return; 21 int mid((l+r)>>1); 22 build(lch[x],l,mid); 23 build(rch[x],mid+1,r); 24 } 25 inline void update(int &x,int las,int pos,int l,int r){ 26 x=++cnt; 27 lch[x]=lch[las]; 28 rch[x]=rch[las]; 29 sum[x]=sum[las]+1; 30 if(l==r) 31 return; 32 int mid((l+r)>>1); 33 if(pos<=mid) 34 update(lch[x],lch[las],pos,l,mid); 35 else 36 update(rch[x],rch[las],pos,mid+1,r); 37 } 38 inline int query(int x,int y,int l,int r,int k){ 39 if(l==r){ 40 if(sum[y]-sum[x]<=k) 41 return 0; 42 return l; 43 } 44 int mid((l+r)>>1); 45 if(sum[lch[y]]-sum[lch[x]]>k) 46 return query(lch[x],lch[y],l,mid,k); 47 if(sum[rch[y]]-sum[rch[x]]>k) 48 return query(rch[x],rch[y],mid+1,r,k); 49 return 0; 50 } 51 int main(){ 52 n=read(),m=read(); 53 for(int i=1;i<=n;++i) 54 a[i]=read(); 55 build(rt[0],1,n); 56 for(int i=1;i<=n;++i) 57 update(rt[i],rt[i-1],a[i],1,n); 58 while(m--){ 59 int l(read()),r(read()); 60 printf("%d\n",query(rt[l-1],rt[r],1,n,(r-l+1)>>1)); 61 } 62 }
[BZOJ3653]谈笑风生
这道题没来得及写博客，以深度建树，维护$size$的和，$x$，$y$的答案=$size[x]\times min(deep[x],y)+dfs$序在$l[x]$到$r[x]$之间且深度在$deep[x]+1$到$deep[x]+k$之间的$size$和（话说这道题扔这个板块合适吗= =）
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 using namespace std; 6 inline int read(){ 7 int sum(0); 8 char ch(getchar()); 9 for(;ch<'0'||ch>'9';ch=getchar()); 10 for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 11 return sum; 12 } 13 struct edge{ 14 int e; 15 edge *n; 16 }a[600005],*pre[300005]; 17 int tot; 18 inline void insert(int s,int e){ 19 a[++tot].e=e; 20 a[tot].n=pre[s]; 21 pre[s]=&a[tot]; 22 } 23 typedef long long L; 24 int n,q; 25 int timee; 26 int l[300005],r[300005],pos[300005],size[300005],fa[300005],dep[300005]; 27 inline void dfs(int u){ 28 l[u]=++timee; 29 pos[timee]=u; 30 size[u]=1; 31 for(edge *i=pre[u];i;i=i->n){ 32 int e(i->e); 33 if(e!=fa[u]){ 34 fa[e]=u; 35 dep[e]=dep[u]+1; 36 dfs(e); 37 size[u]+=size[e]; 38 } 39 } 40 r[u]=timee; 41 } 42 int mx,cnt; 43 int rt[300005],lch[6000005],rch[6000005]; 44 L sum[6000005]; 45 inline void update(int &x,int las,int pos,int w,int l,int r){ 46 x=++cnt; 47 lch[x]=lch[las]; 48 rch[x]=rch[las]; 49 sum[x]=sum[las]+w; 50 if(l==r) 51 return; 52 int mid((l+r)>>1); 53 if(pos<=mid) 54 update(lch[x],lch[las],pos,w,l,mid); 55 else 56 update(rch[x],rch[las],pos,w,mid+1,r); 57 } 58 inline L query(int x,int y,int ll,int rr,int l,int r){ 59 if(ll<=l&&r<=rr) 60 return sum[y]-sum[x]; 61 int mid((l+r)>>1); 62 L ret(0); 63 if(ll<=mid) 64 ret+=query(lch[x],lch[y],ll,rr,l,mid); 65 if(mid<rr) 66 ret+=query(rch[x],rch[y],ll,rr,mid+1,r); 67 return ret; 68 } 69 inline int gg(){ 70 freopen("laugh.in","r",stdin); 71 freopen("laugh.out","w",stdout); 72 n=read(),q=read(); 73 for(int i=1;i<n;++i){ 74 int x(read()),y(read()); 75 insert(x,y),insert(y,x); 76 } 77 dfs(1); 78 for(int i=1;i<=n;++i) 79 mx=max(dep[i],mx); 80 for(int i=1;i<=n;++i) 81 update(rt[i],rt[i-1],dep[pos[i]],size[pos[i]]-1,0,mx); 82 while(q--){ 83 int x(read()),y(read()); 84 L ans(0); 85 ans+=(L)(size[x]-1)*(L)min(dep[x],y); 86 ans+=query(rt[l[x]],rt[r[x]],dep[x]+1,dep[x]+y,0,mx); 87 printf("%lld\n",ans); 88 } 89 } 90 int K(gg()); 91 int main(){;}
动态主席树（树状数组套主席树）
实际上没有啥，就是把维护前缀和的工具改为的树状数组式的而已，具体的只要理解的主席树与树状数组，其余的就很好打了
板子题
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 using namespace std; 6 inline int read(){ 7 int sum(0); 8 char ch(getchar()); 9 for(;ch<'0'||ch>'9';ch=getchar()); 10 for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 11 return sum; 12 } 13 int n,m; 14 int tot,size,cnt; 15 int v[10005],num[20005],has[20005]; 16 int rt[10005],sum[2500005],lch[2500005],rch[2500005]; 17 int A[10005],B[10005],K[10005]; 18 bool flag[10005]; 19 int a,b,L[30],R[30]; 20 char op[2]; 21 inline int lowbit(int x){ 22 return x&-x; 23 } 24 inline int find(int x){ 25 int l(1),r(size),mid; 26 while(l<=r){ 27 mid=(l+r)>>1; 28 if(has[mid]<x) 29 l=mid+1; 30 else 31 r=mid-1; 32 } 33 return l; 34 //return lower_bound(has+1,has+size+1,x)-has; 35 } 36 inline void update(int &x,int las,int pos,int w,int l,int r){ 37 x=++cnt; 38 lch[x]=lch[las]; 39 rch[x]=rch[las]; 40 sum[x]=sum[las]+w; 41 if(l==r) 42 return; 43 int mid((l+r)>>1); 44 if(pos<=mid) 45 update(lch[x],lch[las],pos,w,l,mid); 46 else 47 update(rch[x],rch[las],pos,w,mid+1,r); 48 } 49 inline int query(int l,int r,int k){ 50 if(l==r) 51 return l; 52 int suml(0),sumr(0),mid((l+r)>>1); 53 for(int i=1;i<=a;++i)suml+=sum[lch[L[i]]]; 54 for(int i=1;i<=b;++i)sumr+=sum[lch[R[i]]]; 55 if(sumr-suml>=k){ 56 for(int i=1;i<=a;++i)L[i]=lch[L[i]]; 57 for(int i=1;i<=b;++i)R[i]=lch[R[i]]; 58 return query(l,mid,k); 59 } 60 else{ 61 for(int i=1;i<=a;++i)L[i]=rch[L[i]]; 62 for(int i=1;i<=b;++i)R[i]=rch[R[i]]; 63 return query(mid+1,r,k-(sumr-suml)); 64 } 65 } 66 int main(){ 67 n=read(),m=read(); 68 for(int i=1;i<=n;++i){ 69 v[i]=read(); 70 num[++tot]=v[i]; 71 } 72 for(int i=1;i<=m;++i){ 73 scanf("%s",op); 74 if(op[0]=='Q'){ 75 flag[i]=1; 76 A[i]=read(); 77 B[i]=read(); 78 K[i]=read(); 79 } 80 else{ 81 A[i]=read(); 82 B[i]=read(); 83 num[++tot]=B[i]; 84 } 85 } 86 sort(num+1,num+tot+1); 87 has[++size]=num[1]; 88 for(int i=2;i<=tot;++i) 89 if(num[i]!=num[i-1]) 90 has[++size]=num[i]; 91 for(int i=1;i<=n;++i){ 92 int tmp(find(v[i])); 93 for(int j=i;j<=n;j+=lowbit(j)) 94 update(rt[j],rt[j],tmp,1,1,size); 95 } 96 for(int i=1;i<=m;++i){ 97 if(flag[i]){ 98 a=b=0; 99 --A[i]; 100 for(int j=A[i];j>0;j-=lowbit(j)) 101 L[++a]=rt[j]; 102 for(int j=B[i];j>0;j-=lowbit(j)) 103 R[++b]=rt[j]; 104 printf("%d\n",has[query(1,size,K[i])]); 105 } 106 else{ 107 int tmp(find(v[A[i]])); 108 for(int j=A[i];j<=n;j+=lowbit(j)) 109 update(rt[j],rt[j],tmp,-1,1,size); 110 v[A[i]]=B[i]; 111 tmp=find(B[i]); 112 for(int j=A[i];j<=n;j+=lowbit(j)) 113 update(rt[j],rt[j],tmp,1,1,size); 114 } 115 } 116 }
[CQOI2015]任务查询系统
其实是静态的，但是要用树状数组套一下，所以就扔在这里了
差分一下，瞎XX乱搞就行了
1 #include<algorithm> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdio> 5 using namespace std; 6 typedef long long L; 7 inline L read(){ 8 L sum(0),f(1); 9 char ch(getchar()); 10 for(;ch<'0'||ch>'9';ch=getchar()) 11 if(ch=='-') 12 f=-1; 13 for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 14 return sum*f; 15 } 16 int m,n; 17 int cnt,tot; 18 int q[100005],top; 19 int s[100005],e[100005],p[100005],num[100005]; 20 int rt[100005],size[20000005],lch[20000005],rch[20000005]; 21 L sum[20000005]; 22 L ans(1),x,a,b,c,k; 23 inline int lowbit(int x){ 24 return x&-x; 25 } 26 inline int insert(int x,int val,int f){ 27 if(!x) 28 x=++cnt; 29 size[x]+=f; 30 sum[x]+=f*num[val]; 31 int tmp(x); 32 int l(1),r(m); 33 while(l<r){ 34 int mid((l+r)>>1); 35 if(val<=mid){ 36 r=mid; 37 if(!lch[x]) 38 lch[x]=++cnt; 39 x=lch[x]; 40 size[x]+=f; 41 sum[x]+=f*num[val]; 42 } 43 else{ 44 l=mid+1; 45 if(!rch[x]) 46 rch[x]=++cnt; 47 x=rch[x]; 48 size[x]+=f; 49 sum[x]+=f*num[val]; 50 } 51 } 52 return tmp; 53 } 54 inline void update(int x,int val,int f){ 55 for(int i=x;i<=n;i+=lowbit(i)) 56 rt[i]=insert(rt[i],val,f); 57 } 58 inline L query(int x,int k){ 59 L ret(0),tot(0); 60 top=0; 61 for(int i=x;i>0;i-=lowbit(i)){ 62 q[++top]=rt[i]; 63 tot+=size[rt[i]]; 64 ret+=sum[rt[i]]; 65 } 66 if(k>tot) 67 return ret; 68 ret=0; 69 int l(1),r(m); 70 while(l<r){ 71 int mid((l+r)>>1),tmp(0); 72 for(int i=1;i<=top;++i) 73 tmp+=size[lch[q[i]]]; 74 if(tmp<k){ 75 for(int i=1;i<=top;++i) 76 ret+=sum[lch[q[i]]]; 77 k-=tmp; 78 l=mid+1; 79 for(int i=1;i<=top;++i) 80 q[i]=rch[q[i]]; 81 } 82 else{ 83 r=mid; 84 for(int i=1;i<=top;++i) 85 q[i]=lch[q[i]]; 86 } 87 } 88 ret+=1LL*k*num[l]; 89 return ret; 90 } 91 int main(){ 92 m=read(),n=read(); 93 for(int i=1;i<=m;++i) 94 s[i]=read(),e[i]=read(),num[i]=p[i]=read(); 95 sort(num+1,num+m+1); 96 tot=unique(num+1,num+m+1)-num-1; 97 for(int i=1;i<=m;++i){ 98 p[i]=lower_bound(num+1,num+tot+1,p[i])-num; 99 update(s[i],p[i],1); 100 update(e[i]+1,p[i],-1); 101 } 102 for(int i=1;i<=n;++i){ 103 x=read(),a=read(),b=read(),c=read(); 104 k=1+(a*ans+b)%c; 105 ans=query(x,k); 106 printf("%lld\n",ans); 107 } 108 }
[BZOJ 4999]This Problem Is Too Simple!
树剖上乱搞
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<map> 5 using namespace std; 6 inline int read(){ 7 int sum(0); 8 char ch(getchar()); 9 for(;ch<'0'||ch>'9';ch=getchar()); 10 for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 11 return sum; 12 } 13 struct edge{ 14 int e; 15 edge *n; 16 }ed[200005],*pre[100005]; 17 int tot; 18 inline void insert(int s,int e){ 19 ed[++tot].e=e; 20 ed[tot].n=pre[s]; 21 pre[s]=&ed[tot]; 22 } 23 map<int,int>ma; 24 int num; 25 int n,q; 26 int a[100005]; 27 int dep[100005],size[100005],fa[100005],son[100005]; 28 inline void dfs1(int u){ 29 size[u]=1; 30 son[u]=0; 31 for(edge *i=pre[u];i;i=i->n){ 32 int e(i->e); 33 if(e!=fa[u]){ 34 fa[e]=u; 35 dep[e]=dep[u]+1; 36 dfs1(e); 37 size[u]+=size[e]; 38 if(size[e]>size[son[u]]) 39 son[u]=e; 40 } 41 } 42 } 43 int timee; 44 int id[100005],pos[100005],top[100005]; 45 inline void dfs2(int u,int rt){ 46 top[u]=rt; 47 id[u]=++timee; 48 pos[timee]=u; 49 if(son[u]) 50 dfs2(son[u],rt); 51 for(edge *i=pre[u];i;i=i->n){ 52 int e(i->e); 53 if(e!=fa[u]&&e!=son[u]) 54 dfs2(e,e); 55 } 56 } 57 int cnt; 58 int rt[300005],lch[12000005],rch[12000005],sum[12000005]; 59 inline void update(int &x,int pos,int w,int l,int r){ 60 if(!x) 61 x=++cnt; 62 sum[x]+=w; 63 if(l==r) 64 return; 65 int mid((l+r)>>1); 66 if(pos<=mid) 67 update(lch[x],pos,w,l,mid); 68 else 69 update(rch[x],pos,w,mid+1,r); 70 } 71 inline int query(int x,int ll,int rr,int l,int r){ 72 if(!x) 73 return 0; 74 if(ll<=l&&r<=rr) 75 return sum[x]; 76 int mid((l+r)>>1),ret(0); 77 if(ll<=mid) 78 ret+=query(lch[x],ll,rr,l,mid); 79 if(mid<rr) 80 ret+=query(rch[x],ll,rr,mid+1,r); 81 return ret; 82 } 83 inline int ask(int x,int y,int z){ 84 int ret(0),tmp(ma[z]); 85 while(top[x]^top[y]){ 86 if(dep[top[x]]<dep[top[y]]) 87 swap(x,y); 88 ret+=query(rt[tmp],id[top[x]],id[x],1,n); 89 x=fa[top[x]]; 90 } 91 if(dep[x]>dep[y]) 92 swap(x,y); 93 ret+=query(rt[tmp],id[x],id[y],1,n); 94 return ret; 95 } 96 char op[2]; 97 int main(){ 98 memset(pre,NULL,sizeof(pre)); 99 n=read(),q=read(); 100 for(int i=1;i<=n;++i) 101 a[i]=read(); 102 for(int i=1;i<n;++i){ 103 int x(read()),y(read()); 104 insert(x,y),insert(y,x); 105 } 106 dfs1(1); 107 dfs2(1,1); 108 for(int i=1;i<=n;++i){ 109 if(!ma[a[i]]) 110 ma[a[i]]=++num; 111 update(rt[ma[a[i]]],id[i],1,1,n); 112 } 113 while(q--){ 114 scanf("%s",op); 115 if(op[0]=='C'){ 116 int x(read()),y(read()); 117 update(rt[ma[a[x]]],id[x],-1,1,n); 118 if(!ma[y]) 119 ma[y]=++num; 120 a[x]=y; 121 update(rt[ma[y]],id[x],1,1,n); 122 } 123 else{ 124 int x(read()),y(read()),z(read()); 125 if(!ma[z]) 126 puts("0"); 127 else 128 printf("%d\n",ask(x,y,z)); 129 } 130 } 131 }
突然发现根本没做几道题，我好弱啊QAQ
可。。。可。。。可持久化？
真正的可持久化到来，有关历史版本的修改与查询，实际上理解了主席树的修改操作，这个就变得异常的简单起来
[COGS 2554][福利]可持久化线段树
单点修改的板子题，没啥好说的
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 inline int read(){ 6 int sum(0); 7 char ch(getchar()); 8 for(;ch<'0'||ch>'9';ch=getchar()); 9 for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 10 return sum; 11 } 12 int n,q,t(1); 13 int a[10005]; 14 int cnt; 15 int rt[100005],lch[20000005],rch[20000005],mx[20000005]; 16 inline void pushup(int i){ 17 mx[i]=max(mx[lch[i]],mx[rch[i]]); 18 } 19 inline void build(int &x,int l,int r){ 20 x=++cnt; 21 if(l==r){ 22 mx[x]=a[l]; 23 return; 24 } 25 int mid((l+r)>>1); 26 build(lch[x],l,mid); 27 build(rch[x],mid+1,r); 28 pushup(x); 29 } 30 inline void update(int &x,int las,int pos,int w,int l,int r){ 31 x=++cnt; 32 lch[x]=lch[las]; 33 rch[x]=rch[las]; 34 mx[x]=mx[las]; 35 if(l==r){ 36 mx[x]=w; 37 return; 38 } 39 int mid((l+r)>>1); 40 if(pos<=mid) 41 update(lch[x],lch[las],pos,w,l,mid); 42 else 43 update(rch[x],rch[las],pos,w,mid+1,r); 44 pushup(x); 45 } 46 inline int query(int x,int ll,int rr,int l,int r){ 47 if(ll==l&&r==rr) 48 return mx[x]; 49 int mid((l+r)>>1); 50 if(rr<=mid) 51 return query(lch[x],ll,rr,l,mid); 52 if(mid<ll) 53 return query(rch[x],ll,rr,mid+1,r); 54 return max(query(lch[x],ll,mid,l,mid),query(rch[x],mid+1,rr,mid+1,r)); 55 } 56 inline int gg(){ 57 freopen("longterm_segtree.in","r",stdin); 58 freopen("longterm_segtree.out","w",stdout); 59 n=read(),q=read(); 60 for(int i=1;i<=n;++i) 61 a[i]=read(); 62 build(rt[1],1,n); 63 while(q--){ 64 int op(read()),x(read()),y(read()),z(read()); 65 if(op==0) 66 printf("%d\n",query(rt[x],y,z,1,n)); 67 else 68 update(rt[++t],rt[x],y,z,1,n); 69 } 70 return 0; 71 } 72 int K(gg()); 73 int main(){;}
可持久化线段树区间修改的板子题
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 typedef long long L; 6 inline int read(){ 7 int sum(0),f(1); 8 char ch(getchar()); 9 for(;ch<'0'||ch>'9';ch=getchar()) 10 if(ch=='-') 11 f=-1; 12 for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 13 return sum*f; 14 } 15 int n,m,t,cnt; 16 int v[100005]; 17 int rt[100005],lch[4000005],rch[4000005]; 18 L sum[4000005],add[4000005]; 19 char op[2]; 20 inline int build(int l,int r){ 21 int x(++cnt); 22 add[x]=0; 23 if(l==r){ 24 sum[x]=v[l]; 25 lch[x]=rch[x]=0; 26 // cout<<"pos "<<x<<" "<<l<<' '<<r<<" "<<v[l]<<' '<<sum[x]<<endl; 27 return x; 28 } 29 int mid((l+r)>>1); 30 lch[x]=build(l,mid); 31 // cout<<"left child"<<x<<" "<<lch[x]<<' '<<sum[lch[x]]<<endl; 32 rch[x]=build(mid+1,r); 33 // cout<<"right child"<<x<<" "<<rch[x]<<' '<<sum[rch[x]]<<endl; 34 // cout<<"wtf"<<endl; 35 sum[x]=sum[lch[x]]+sum[rch[x]]; 36 // cout<<"wtf"<<endl; 37 // cout<<x<<" "<<l<<' '<<r<<" "<<lch[x]<<' '<<rch[x]<<' '<<sum[lch[x]]<<' '<<sum[rch[x]]<<' '<<sum[x]<<endl; 38 return x; 39 } 40 inline void pushup(int x,int len){ 41 sum[x]=sum[lch[x]]+sum[rch[x]]+add[lch[x]]*(len-(len>>1))+add[rch[x]]*(len>>1); 42 } 43 inline int update(int rt,int ll,int rr,L w,int l,int r){ 44 int newrt(++cnt); 45 add[newrt]=add[rt]; 46 if(ll<=l&&r<=rr){ 47 sum[newrt]=sum[rt]; 48 add[newrt]=add[rt]+w; 49 lch[newrt]=lch[rt]; 50 rch[newrt]=rch[rt]; 51 return newrt; 52 } 53 int mid((l+r)>>1); 54 if(ll<=mid) 55 lch[newrt]=update(lch[rt],ll,rr,w,l,mid); 56 else 57 lch[newrt]=lch[rt]; 58 if(mid<rr) 59 rch[newrt]=update(rch[rt],ll,rr,w,mid+1,r); 60 else 61 rch[newrt]=rch[rt]; 62 pushup(newrt,r-l+1); 63 return newrt; 64 } 65 inline L query(int rt,int ll,int rr,int l,int r,L add1){ 66 // cout<<rt<<' '<<ll<<' '<<rr<<' '<<l<<' '<<r<<' '<<add1<<endl; 67 if(ll<=l&&r<=rr) 68 return sum[rt]+(add1+add[rt])*(r-l+1); 69 int mid((l+r)>>1); 70 L ret(0); 71 if(ll<=mid) 72 ret+=query(lch[rt],ll,rr,l,mid,add[rt]+add1); 73 if(mid<rr) 74 ret+=query(rch[rt],ll,rr,mid+1,r,add[rt]+add1); 75 return ret; 76 } 77 int main(){ 78 int flag(0); 79 while(scanf("%d%d",&n,&m)==2){ 80 if(flag) 81 puts(""); 82 for(int i=1;i<=n;++i) 83 v[i]=read(); 84 ++flag; 85 cnt=t=0; 86 rt[0]=build(1,n); 87 while(m--){ 88 scanf("%s",op); 89 if(op[0]=='Q'){ 90 int x(read()),y(read()); 91 printf("%lld\n",query(rt[t],x,y,1,n,0)); 92 } 93 if(op[0]=='C'){ 94 int x(read()),y(read()); 95 L z(read()); 96 rt[t+1]=update(rt[t],x,y,z,1,n); 97 ++t; 98 } 99 if(op[0]=='H'){ 100 int x(read()),y(read()),z(read()); 101 printf("%lld\n",query(rt[z],x,y,1,n,0)); 102 } 103 if(op[0]=='B'){ 104 int x(read()); 105 t=x; 106 cnt=rt[t+1]-1; 107 } 108 } 109 } 110 }
[BZOJ 3221]Obserbing the tree树上询问
真正做过的难题出现！
树剖+可持久化线段树区间修改+线段树上维护等差数列（其实没多难，我太弱而已= =）
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 typedef long long L; 6 inline L read(){ 7 L sum(0),f(1); 8 char ch(getchar()); 9 for(;ch<'0'||ch>'9';ch=getchar()) 10 if(ch=='-') 11 f=-1; 12 for(;ch>='0'&&ch<='9';sum=sum*10+(ch^48),ch=getchar()); 13 return sum*f; 14 } 15 struct edge{ 16 int e; 17 edge *n; 18 }a[200005],*pre[100005]; 19 int to; 20 inline void insert(int s,int e){ 21 a[++to].e=e; 22 a[to].n=pre[s]; 23 pre[s]=&a[to]; 24 } 25 int n,m,now,tot; 26 char op[2]; 27 int dep[100005],size[100005],fa[100005],son[100005]; 28 inline void dfs1(int u){ 29 size[u]=1; 30 son[u]=0; 31 dep[u]=dep[fa[u]]+1; 32 for(edge *i=pre[u];i;i=i->n){ 33 int e(i->e); 34 if(e!=fa[u]){ 35 fa[e]=u; 36 dfs1(e); 37 size[u]+=size[e]; 38 if(size[e]>size[son[u]]) 39 son[u]=e; 40 } 41 } 42 } 43 int timee; 44 int id[100005],top[100005]; 45 inline void dfs2(int u,int rt){ 46 top[u]=rt; 47 id[u]=++timee; 48 if(son[u]) 49 dfs2(son[u],rt); 50 for(edge *i=pre[u];i;i=i->n){ 51 int e(i->e); 52 if(e!=son[u]&&e!=fa[u]) 53 dfs2(e,e); 54 } 55 } 56 inline int lca(int x,int y){ 57 while(top[x]^top[y]){ 58 if(dep[top[x]]<dep[top[y]]) 59 swap(x,y); 60 x=fa[top[x]]; 61 } 62 return dep[x]<dep[y]?x:y; 63 } 64 int rt[100005],lch[20000005],rch[20000005]; 65 int cnt; 66 L sum[20000005],add_sx[20000005],add_gc[20000005]; 67 L ans; 68 inline void update(int &x,int las,int ll,int rr,L sx,L gc,int l,int r){ 69 x=++cnt; 70 lch[x]=lch[las]; 71 rch[x]=rch[las]; 72 sum[x]=sum[las]; 73 add_sx[x]=add_sx[las]; 74 add_gc[x]=add_gc[las]; 75 if(ll==l&&r==rr){ 76 add_sx[x]+=sx; 77 add_gc[x]+=gc; 78 return; 79 } 80 sum[x]+=(sx+sx+gc*(L)(rr-ll))*(rr-ll+1)/2; 81 int mid((l+r)>>1); 82 if(rr<=mid) 83 update(lch[x],lch[las],ll,rr,sx,gc,l,mid); 84 else 85 if(mid<ll) 86 update(rch[x],rch[las],ll,rr,sx,gc,mid+1,r); 87 else{ 88 update(lch[x],lch[las],ll,mid,sx,gc,l,mid); 89 update(rch[x],rch[las],mid+1,rr,sx+(mid-ll+1)*gc,gc,mid+1,r); 90 } 91 } 92 inline L query(int x,int ll,int rr,int l,int r){ 93 L ret((add_sx[x]+(ll-l)*add_gc[x]+add_sx[x]+(rr-l)*add_gc[x])*(rr-ll+1)/2); 94 if(ll==l&&r==rr) 95 return ret+sum[x]; 96 int mid((l+r)>>1); 97 if(rr<=mid) 98 return ret+query(lch[x],ll,rr,l,mid); 99 if(mid<ll) 100 return ret+=query(rch[x],ll,rr,mid+1,r); 101 return ret+query(lch[x],ll,mid,l,mid)+query(rch[x],mid+1,rr,mid+1,r); 102 } 103 inline void change(int x,int y,L sx,L gc){ 104 int f(lca(x,y)); 105 L tp1(0),tp2(dep[x]+dep[y]-dep[f]*2+2); 106 while(top[x]^top[f]){ 107 tp1+=dep[x]-dep[top[x]]+1; 108 update(rt[now],rt[now],id[top[x]],id[x],sx+(tp1-1)*gc,-gc,1,n); 109 x=fa[top[x]]; 110 } 111 while(top[y]^top[f]){ 112 tp2-=dep[y]-dep[top[y]]+1; 113 update(rt[now],rt[now],id[top[y]],id[y],sx+(tp2-1)*gc,gc,1,n); 114 y=fa[top[y]]; 115 } 116 ++tp1; 117 --tp2; 118 if(dep[x]<dep[y]) 119 update(rt[now],rt[now],id[x],id[y],sx+(tp1-1)*gc,gc,1,n); 120 else 121 update(rt[now],rt[now],id[y],id[x],sx+(tp2-1)*gc,-gc,1,n); 122 } 123 inline L ask(int x,int y){ 124 L ret(0); 125 while(top[x]^top[y]){ 126 if(dep[top[x]]<dep[top[y]]) 127 swap(x,y); 128 ret+=query(rt[now],id[top[x]],id[x],1,n); 129 x=fa[top[x]]; 130 } 131 if(dep[x]>dep[y]) 132 swap(x,y); 133 ret+=query(rt[now],id[x],id[y],1,n); 134 return ret; 135 } 136 int main(){ 137 memset(pre,NULL,sizeof(pre)); 138 n=read(),m=read(); 139 for(int i=1;i<n;++i){ 140 int x(read()),y(read()); 141 insert(x,y),insert(y,x); 142 } 143 dfs1(1); 144 dfs2(1,1); 145 while(m--){ 146 scanf("%s",op); 147 if(op[0]=='c'){ 148 L x(read()),y(read()),sx(read()),gc(read()); 149 x^=ans,y^=ans; 150 rt[++tot]=rt[now]; 151 now=tot; 152 change(x,y,sx,gc); 153 } 154 if(op[0]=='q'){ 155 L x(read()),y(read()); 156 x^=ans,y^=ans; 157 ans=ask(x,y); 158 printf("%lld\n",ans); 159 } 160 if(op[0]=='l'){ 161 L x(read()); 162 x^=ans; 163 now=x; 164 } 165 } 166 }
REAL 小结？
反正主席树这东西用处不少，就是联赛貌似用不到，所以只能期望自己能撑过联赛，自己也要加油啊
