树刨的梗来自于Yveh在SDWC2018上的课件
线段树背今晚的大锅
在这里再说一点轶事,据说集训队大爷WZD神犇出过一道带八个操作的LCT
Exciting.
还有我成功拿到了4k的长度,代码更加高清了。
//牺牲一些长度换易读性吧 反正手速快。[滑稽
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#include<cstdio> #include<iostream> #include<cmath> #include<cstdlib> #include<cstring> #include<string> #include<iterator> #include<algorithm> #include<queue> #include<vector> using namespace std; struct Segment_Tree { int l,r,lazy_tag,w; }SGT[400010]={0}; struct ed { int pre,to; }edge[200010]={0}; int nodenum,rootx,opt,modx,fx,tx,at=1,pointer[100010]={0}; int nodev[100010]={0},lx,rx,val,oopt; int nnv[100010]={0},topx[100010]={0},dfn[100010]={0}; int hson[100010]={0},tsize[100010]={0},depth[100010]={0},anc[100010]={0}; int tstamp=0; inline void readx(int& x) { x=0; int k=1; register char ch=0; while (ch<'0' || ch>'9') { ch=getchar(); if (ch=='-') k=-1; } while (ch>='0' && ch<='9') { x=x*10+ch-'0'; ch=getchar(); } x*=k; } inline void Insert() { edge[at].pre=pointer[fx]; edge[at].to=tx; pointer[fx]=at; at++; edge[at].pre=pointer[tx]; edge[at].to=fx; pointer[tx]=at; at++; } inline void Fancy(int nownode,int nowdep,int fa) { depth[nownode]=nowdep; tsize[nownode]=1; anc[nownode]=fa; int prex=pointer[nownode]; while (prex) { if (edge[prex].to!=fa) { Fancy(edge[prex].to,nowdep+1,nownode); tsize[nownode]+=tsize[edge[prex].to]; if (tsize[edge[prex].to]>tsize[hson[nownode]]) hson[nownode]=edge[prex].to; } prex=edge[prex].pre; } } inline void Dreams(int nownode,int src) { dfn[nownode]=++tstamp; topx[nownode]=src; nnv[tstamp]=nodev[nownode]%modx; if (hson[nownode]) Dreams(hson[nownode],src); int prex=pointer[nownode]; while (prex) { if (edge[prex].to!=anc[nownode] && edge[prex].to!=hson[nownode]) Dreams(edge[prex].to,edge[prex].to); prex=edge[prex].pre; } } inline void BuildTree(int leftx,int rightx,int inx) { SGT[inx].l=leftx; SGT[inx].r=rightx; if (leftx==rightx) { SGT[inx].w=nnv[leftx]; return; } int mid=(leftx+rightx)>>1; BuildTree(leftx,mid,inx<<1); BuildTree(mid+1,rightx,inx<<1|1); SGT[inx].w=(SGT[inx<<1].w+SGT[inx<<1|1].w)%modx; } inline void pushdown(int inx) { SGT[inx<<1].lazy_tag+=SGT[inx].lazy_tag; SGT[inx<<1].lazy_tag%=modx; SGT[inx<<1|1].lazy_tag+=SGT[inx].lazy_tag; SGT[inx<<1|1].lazy_tag%=modx; SGT[inx<<1].w+=(SGT[inx<<1].r-SGT[inx<<1].l+1)*SGT[inx].lazy_tag; SGT[inx<<1].w%=modx; SGT[inx<<1|1].w+=(SGT[inx<<1|1].r-SGT[inx<<1|1].l+1)*SGT[inx].lazy_tag; SGT[inx<<1|1].w%=modx; SGT[inx].lazy_tag=0; } inline void update(int inx) { if (SGT[inx].l>=lx && SGT[inx].r<=rx) { SGT[inx].w+=val*(SGT[inx].r-SGT[inx].l+1); SGT[inx].w%=modx; SGT[inx].lazy_tag+=val; SGT[inx].lazy_tag%=modx; return; } if (SGT[inx].lazy_tag) pushdown(inx); int mid=(SGT[inx].l+SGT[inx].r)>>1; if (lx<=mid) update(inx<<1); if (rx>mid) update(inx<<1|1); SGT[inx].w=(SGT[inx<<1].w+SGT[inx<<1|1].w)%modx; } inline int query(int inx) { if (SGT[inx].l>=lx && SGT[inx].r<=rx) return SGT[inx].w; if (SGT[inx].lazy_tag) pushdown(inx); int mid=(SGT[inx].l+SGT[inx].r)>>1,ret=0; if (mid>=lx) ret=(ret+query(inx<<1))%modx; if (mid<rx) ret=(ret+query(inx<<1|1))%modx; return ret; } inline int solve_range(int u,int v) { int ans=0; while (topx[u]!=topx[v]) { if (depth[topx[u]]<depth[topx[v]]) swap(u,v); lx=dfn[topx[u]]; rx=dfn[u]; ans+=query(1); ans%=modx; u=anc[topx[u]]; } if (depth[u]<depth[v]) swap(u,v); lx=dfn[v]; rx=dfn[u]; ans+=query(1); return ans%modx; } inline void upd_range(int u,int v) { while (topx[u]!=topx[v]) { if (depth[topx[u]]<depth[topx[v]]) swap(u,v); lx=dfn[topx[u]]; rx=dfn[u]; update(1); u=anc[topx[u]]; } if (depth[u]<depth[v]) swap(u,v); lx=dfn[v]; rx=dfn[u]; update(1); } inline int solve_subt(int u) { lx=dfn[u]; rx=dfn[u]+tsize[u]-1; int ans=query(1); return ans%modx; } inline void upd_subt(int u) { lx=dfn[u]; rx=dfn[u]+tsize[u]-1; update(1); } int main() { readx(nodenum); readx(opt); readx(rootx); readx(modx); for (int i=1;i<=nodenum;i++) readx(nodev[i]); for (int i=1;i<nodenum;i++) { readx(fx); readx(tx); Insert(); } Fancy(rootx,1,-1); Dreams(rootx,rootx); BuildTree(1,tstamp,1); int ux,vx; for (int i=1;i<=opt;i++) { readx(oopt); if (oopt==1) { readx(ux); readx(vx); readx(val); upd_range(ux,vx); } else if (oopt==2) { readx(ux); readx(vx); printf("%d\n",solve_range(ux,vx)); } else if (oopt==3) { readx(ux); readx(val); upd_subt(ux); } else { readx(ux); printf("%d\n",solve_subt(ux)); } } return 0; } |
没错这就是洛谷的板子题.