【树剖求LCA】树剖知识点

不太优美但是有注释的版本：

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#include<cstdio> #include<iostream> using namespace std; struct edge{     int to,ne; }e[1000005]; int n,m,s,ecnt,head[500005],dep[500005],siz[500005],son[500005],top[500005],f[500005]; void add(int x,int y)    //加边 {     e[++ecnt].to=y;     e[ecnt].ne=head[x];     head[x]=ecnt; } void dfs1(int x)                             //构造树 {     siz[x]=1;                                //假设当前节点仅有一个儿子     dep[x]=dep[f[x]]+1;                      //当前节点深度=父亲节点深度+1     for(int i=head[x];i;i=e[i].ne)          //遍历所有的子节点     {         int dd=e[i].to;         if(dd==f[x])continue;               //如果是父节点,则略过         f[dd]=x;                             //那么确定x是当前节点的父亲         dfs1(dd);                            //向下遍历         siz[x]+=siz[dd];                     //遍历完子树之后,加上子树的大小         if(!son[x]||siz[son[x]]<siz[dd])    //如果x节点重儿子未确定或者重儿子的子树比当前遍历节点的子树小             son[x]=dd;                       //更新重儿子     } }   void dfs2(int x,int tv)                     //求重链 {     top[x]=tv;                               //设置x所在重链顶为tv     if(son[x])dfs2(son[x],tv);              //如果x有重儿子,那么随着这条重链走     for(int i=head[x];i;i=e[i].ne)                {         int dd=e[i].to;         if(dd==f[x]||dd==son[x])continue;  //如果走到父亲或者走到重儿子(已经走过重儿子,避免重复),那么跳过         dfs2(dd,dd);                        //开启一条新链,链顶是其本身     } } int lca(int x,int y) {     while(top[x]!=top[y])                              //如果二者不在同一条重链上     {         if(dep[top[x]] >= dep[top[y]]) x=f[top[x]];   //选择所在重链的顶的深度较大的点向上跳,目的是防止跳过LCA         else y=f[top[y]];     }     return dep[x] < dep[y] ?x :y;                     //当二者在同一条重链上的时候,选择深度较浅的点即为lca             } int main() {     scanf("%d%d%d",&n,&m,&s);     for(int i=1;i<n;++i)     {         int x,y;         scanf("%d%d",&x,&y);         add(x,y);         add(y,x);     }     dfs1(s);     dfs2(s,s);     for(int i=1;i<=m;++i)     {         int x,y;         scanf("%d%d",&x,&y);         printf("%d\n",lca(x,y));     } }

　　

 

比较优美但是没注释的版本：

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#include<iostream> #include<vector> #include<cstdio> #include<queue> #include<map> #include<cstdlib> #include<cmath> #include<algorithm> #include<set> #include<cstring> using namespace std; typedef long long ll; const ll INF=99999999; const int N = 500010;   int n,m,s;   struct edge{     int to,ne;       }e[N*2];   int top[N],siz[N],son[N],fa[N],dep[N]; int head[N],ecnt = 1;   void add(int x,int y) {     e[ecnt].to = y;     e[ecnt].ne = head[x];           head[x] = ecnt++; }   void dfs1(int x) {     siz[x] = 1;     dep[x] = dep[fa[x]] + 1;           for(int i = head[x];i;i = e[i].ne){         int t = e[i].to;         if(t == fa[x]) continue;         fa[t] = x;                   dfs1(t);         siz[x] += siz[t];         if(!son[x]||siz[son[x]] < siz[t])             son[x] = t;     } }   void dfs2(int x,int tp) {     top[x] = tp;           if(son[x])         dfs2(son[x],tp);               for(int i = head[x];i;i = e[i].ne){         int t = e[i].to;         if(t == son[x]||t == fa[x]) continue;                   dfs2(t,t);     }     }   int lca(int x,int y) {     while(top[x] != top[y]){         if(dep[top[x]] >= dep[top[y]]) x = fa[top[x]];         else y = fa[top[y]];     }     return dep[x] < dep[y] ?x :y; } int main() {           scanf("%d%d%d",&n,&m,&s);     for(int i = 1;i < n;i++){         int x,y;         scanf("%d%d",&x,&y);                   add(x,y);         add(y,x);     }     dfs1(s);     dfs2(s,s);           for(int i = 1;i <= m;i++){         int a,b;         scanf("%d%d",&a,&b);                   printf("%d\n",lca(a,b));     }     return 0; }

 

树剖理解容易，需要注意的是题目如果给的是双向边,e数组需要开两倍于边数