[BZOJ2243][SDOI2011]染色 解题报告|树链剖分

Description

给定一棵有n个节点的无根树和m个操作，操作有2类：

1、将节点a到节点b路径上所有点都染成颜色c；

2、询问节点a到节点b路径上的颜色段数量（连续相同颜色被认为是同一段），如“112221”由3段组成：“11”、“222”和“1”。

请你写一个程序依次完成这m个操作。

　　与上一题差别不大，主要就是solve过程要根据左端点和右端点的颜色处理合并时候的情况

　　线段树的每个节点要记录颜色段数|最左边的颜色|最右边的颜色

　　同时往下传的时候标记要做好（之前那道题是单点修改所以不用考虑下传

　　因为这个昨晚调了一个晚上无果...果然是早上头脑清醒的多 一下子就改好了...

program bzoj2243;
const maxn=;maxm=;
var ter,next:array[-..maxm]of longint;
    link,pos,deep,size,belong,v:array[-..maxn]of longint;
    fa:array[-..maxn,-..]of longint;
    cnt,n,m,i,j,x,y,z,t:longint;
    ch:char;
    tr:array[-..*maxn]of record l,r,lc,rc,cover:longint;wait:boolean;end;

procedure add(x,y:longint);
begin
    inc(j);ter[j]:=y;next[j]:=link[x];link[x]:=j;
    inc(j);ter[j]:=x;next[j]:=link[y];link[y]:=j;
end;

procedure dfs1(p:longint);
var i,j:longint;
begin
    size[p]:=;
    for i:= to  do
    begin
        if deep[p]<= << i then break;
        fa[p][i]:=fa[fa[p][i-]][i-];
    end;
    j:=link[p];
    while j<> do
    begin
        if deep[ter[j]]= then
        begin
            deep[ter[j]]:=deep[p]+;
            fa[ter[j]][]:=p;
            dfs1(ter[j]);
            inc(size[p],size[ter[j]]);
        end;
        j:=next[j];
    end;
end;

procedure dfs2(p,chain:longint);
var j,k:longint;
begin
    inc(cnt);pos[p]:=cnt;belong[p]:=chain;
    k:=;
    j:=link[p];
    while j<> do
    begin
        if deep[ter[j]]>deep[p] then
            if size[ter[j]]>size[k] then k:=ter[j];
        j:=next[j];
    end;
    if k= then exit;
    dfs2(k,chain);
    j:=link[p];
    while j<> do
    begin
        if (deep[ter[j]]>deep[p])and(k<>ter[j]) then dfs2(ter[j],ter[j]);
        j:=next[j];
    end;
end;

procedure build(p,l,r:longint);
var mid:longint;
begin
    tr[p].l:=l;tr[p].r:=r;tr[p].cover:=;tr[p].lc:=-;tr[p].rc:=-;
    if l=r then exit;
    mid:=(l+r) >> ;
    build(p << ,l,mid);
    build(p << +,mid+,r);
end;

procedure insert(p,l,r,x:longint);
var mid:longint;
begin
    if (tr[p].l=l)and(tr[p].r=r) then
    begin
        tr[p].cover:=;tr[p].lc:=x;tr[p].rc:=x;tr[p].wait:=true;
        exit;
    end;
    mid:=(tr[p].l+tr[p].r) >> ;
    if tr[p].wait then
    begin
        tr[p << ].cover:=;tr[p << ].lc:=tr[p].lc;tr[p << ].rc:=tr[p].rc;
        tr[p << +].cover:=;tr[p << +].lc:=tr[p].lc;tr[p << +].rc:=tr[p].rc;
        tr[p].wait:=false;
                tr[p << ].wait:=true;tr[p << +].wait:=true;
    end;
    if r<=mid then insert(p << ,l,r,x) else
        if l>mid then insert(p << +,l,r,x) else
        begin
            insert(p << ,l,mid,x);
            insert(p << +,mid+,r,x);
        end;
        tr[p].cover:=tr[p << ].cover+tr[p << +].cover;
    if tr[p << ].rc=tr[p << +].lc then dec(tr[p].cover);
    tr[p].lc:=tr[p << ].lc;tr[p].rc:=tr[p << +].rc;
end;

function lca(x,y:longint):longint;
var tem,i:longint;
begin
    if deep[x]<deep[y] then
    begin
        tem:=x;x:=y;y:=tem;
    end;
    if deep[x]>deep[y] then
    begin
        i:=trunc(ln(deep[x]-deep[y])/ln());
        while deep[x]<>deep[y] do
        begin
            while (deep[x]-deep[y]>= << i) do x:=fa[x][i];
            dec(i);
        end;
    end;
    if x=y then exit(x);
    i:=trunc(ln(n)/ln());
    while fa[x][]<>fa[y][] do
    begin
        while (fa[x][i]<>fa[y][i]) do
        begin
            x:=fa[x,i];y:=fa[y][i];
        end;
        dec(i);
    end;
    exit(fa[x,]);
end;

procedure query(p,l,r:longint;var lc,rc,ave:longint);
var mid,ll,rr,x,tem:longint;
begin
    if (tr[p].l=l)and(tr[p].r=r) then
    begin
        lc:=tr[p].lc;rc:=tr[p].rc;
        ave:=tr[p].cover;
                exit;
        end;
    mid:=(tr[p].l+tr[p].r) >> ;
        if tr[p].wait then
    begin
        tr[p << ].cover:=;tr[p << ].lc:=tr[p].lc;tr[p << ].rc:=tr[p].rc;
        tr[p << +].cover:=;tr[p << +].lc:=tr[p].lc;tr[p << +].rc:=tr[p].rc;
        tr[p].wait:=false;
                tr[p << ].wait:=true;tr[p << +].wait:=true;
    end;
    if r<=mid then query(p << ,l,r,lc,rc,ave) else
        if l>mid then query(p << +,l,r,lc,rc,ave) else
        begin
            query(p << ,l,mid,lc,ll,ave);
            query(p << +,mid+,r,rr,rc,tem);
                        inc(ave,tem);
            if ll=rr then dec(ave);
        end;
end;

function solve(x,y:longint):longint;
var r,sum,lc,rc,tem:longint;
begin
    r:=-;sum:=;
    while belong[x]<>belong[y] do
    begin
        query(,pos[belong[x]],pos[x],lc,rc,tem);
                inc(sum,tem);
        if rc=r then dec(sum);
        r:=lc;
        x:=fa[belong[x]][];
    end;
    query(,pos[y],pos[x],lc,rc,tem);
        inc(sum,tem);
    if rc=r then dec(sum);
    exit(sum);
end;

procedure mend(x,y,z:longint);
begin
    while belong[x]<>belong[y] do
    begin
        insert(,pos[belong[x]],pos[x],z);
        x:=fa[belong[x]][];
    end;
    insert(,pos[y],pos[x],z);
end;

begin
    readln(n,m);
    for i:= to n do read(v[i]);readln;
    j:=;
    for i:= to n- do
    begin
        readln(x,y);
        add(x,y);
    end;
        deep[]:=;
    dfs1();
    cnt:=;dfs2(,);
        build(,,n);
    for i:= to n do insert(,pos[i],pos[i],v[i]);
        for i:= to m do
    begin
        read(ch);
        if ch='C' then
        begin
            readln(x,y,z);
            t:=lca(x,y);
            mend(x,t,z);mend(y,t,z);
        end else
        begin
            readln(x,y);
                z:=lca(x,y);
                writeln(solve(x,z)+solve(y,z)-);
        end;
    end;
end.

 [UPD.5.11]今天复习树链剖分，挑了这道题写了写。一个地方WA检查了三遍对拍了一个多小时才找出来...

就是在判断链之间答案合并的时候是从右往左的而不是从左往右...习惯性的写法干脆检查的时候都忽略了这一点...

另外这次用C++大概150行左右..感觉还不错

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 #define maxn 1000010
 #define maxm 2000010
 struct node{
     int l,r,lc,rc,count,wait;
 }tr[maxn*];
 int next[maxm],ter[maxm],link[maxn],pos[maxn],belong[maxn],dep[maxn];
 int fa[maxn][],sz[maxn],n,m,e,v[maxn],cnt;
 char ch[];
 void swap(int &x,int &y){
     int tmp=x;x=y;y=tmp;
 }
 void add(int x,int y){
     ter[++e]=y;next[e]=link[x];link[x]=e;
     ter[++e]=x;next[e]=link[y];link[y]=e;
 }
 void dfs1(int p){
     sz[p]=;
     for (int i=;i<=&&dep[p]>(<<i);i++) fa[p][i]=fa[fa[p][i-]][i-];
     for (int i=link[p];i;i=next[i]) if (dep[ter[i]]==){
         dep[ter[i]]=dep[p]+;
         fa[ter[i]][]=p;
         dfs1(ter[i]);
         sz[p]+=sz[ter[i]];
     }
 }
 void dfs2(int p,int chain){
     pos[p]=++cnt;belong[p]=chain;
     int k=;
     for (int i=link[p];i;i=next[i]) if (dep[ter[i]]==dep[p]+&&sz[ter[i]]>sz[k]) k=ter[i];
     if (k==) return;
     dfs2(k,chain);
     for (int i=link[p];i;i=next[i]) if (dep[ter[i]]==dep[p]+&&ter[i]!=k) dfs2(ter[i],ter[i]);
 }
 void update(int p){
     tr[p].lc=tr[p<<].lc;
     tr[p].rc=tr[(p<<)+].rc;
     tr[p].count=tr[p<<].count+tr[(p<<)+].count;
     if (tr[p<<].rc==tr[(p<<)+].lc) tr[p].count--;
 }
 void push(int p){
     if (tr[p].wait==-) return;
     int u=p<<,w=(p<<)+,col=tr[p].wait;
     tr[u].lc=tr[u].rc=tr[u].wait=col;tr[u].count=;
     tr[w].lc=tr[w].rc=tr[w].wait=col;tr[w].count=;
     tr[p].wait=-;
 }
 void build(int p,int l,int r){
     tr[p].l=l;tr[p].r=r;tr[p].wait=-;
     if (l==r) return;
     int mid=(l+r) >> ;
     build(p<<,l,mid);build((p<<)+,mid+,r);
 }
 int lca(int x,int y){
     if (dep[x]<dep[y]) swap(x,y);
     int i;
     if (dep[x]>dep[y]){
         i=(int)(log(dep[x]-dep[y])/log());
         while (dep[x]>dep[y]){
             while (dep[x]-dep[y]>=<<i) x=fa[x][i];
             i--;
         }
     }
     if (x==y) return x;
     i=(int)(log(n)/log());
     while (fa[x][]!=fa[y][]){
         while (fa[x][i]!=fa[y][i]) x=fa[x][i],y=fa[y][i];
         i--;
     }
     return fa[x][];
 }
 void insert(int p,int l,int r,int col){
     if (tr[p].l==l&&tr[p].r==r){
         tr[p].lc=col;tr[p].rc=col;tr[p].wait=col;tr[p].count=;
         return;
     }
     push(p);
     int mid=(tr[p].l+tr[p].r)>>;
     if (r<=mid) insert(p<<,l,r,col);else
         if (l>mid) insert((p<<)+,l,r,col);else{
             insert(p<<,l,mid,col);insert((p<<)+,mid+,r,col);
         }
     update(p);
 }
 void query(int p,int l,int r,int &lc,int &rc,int &ans){
     if (tr[p].l==l&&tr[p].r==r){
         lc=tr[p].lc;rc=tr[p].rc;ans=tr[p].count;
         return;
     }
     push(p);
     int mid=(tr[p].l+tr[p].r)>>;
     if (r<=mid) query(p<<,l,r,lc,rc,ans);else
         if (l>mid) query((p<<)+,l,r,lc,rc,ans);else
         {
             int ll,rr,ans1;
             query(p<<,l,mid,lc,rr,ans);
             query((p<<)+,mid+,r,ll,rc,ans1);
             ans+=ans1;if (ll==rr) ans--;
         }
     update(p);
 }
 void change(int x,int y,int col){
     while (belong[x]!=belong[y]){
         insert(,pos[belong[x]],pos[x],col);
         x=fa[belong[x]][];
     }
     insert(,pos[y],pos[x],col);
 }
 int solve(int x,int y){
     int sum=,r=-,ll,rr,tmp;
     while (belong[x]!=belong[y]){
         query(,pos[belong[x]],pos[x],ll,rr,tmp);
         x=fa[belong[x]][];
         sum+=tmp;if (rr==r) sum--;
         r=ll;
     }
     query(,pos[y],pos[x],ll,rr,tmp);
     sum+=tmp;if (rr==r) sum--;
     return sum;
 }
 int main(){
     scanf("%d%d",&n,&m);
     for (int i=;i<=n;i++) scanf("%d",&v[i]);
     e=;
     int x,y,col,t;
     for (int i=;i<=n-;i++) scanf("%d%d",&x,&y),add(x,y);
     dep[]=;dfs1();
     cnt=;dfs2(,);
     build(,,n);
     for (int i=;i<=n;i++) insert(,pos[i],pos[i],v[i]);
     for (int i=;i<=m;i++){
         scanf("%s",ch);
         if (ch[]=='C'){
             scanf("%d%d%d",&x,&y,&col);
             t=lca(x,y);
             change(x,t,col);change(y,t,col);
         }else{
             scanf("%d%d",&x,&y);
             t=lca(x,y);
             printf("%d\n",solve(x,t)+solve(y,t)-);
         }
     }
     return ;
 }