题目描述
NIO is playing a game about trees.
The game has two trees A,BA, BA,B each with NNN vertices. The vertices in each tree are numbered from 111 to NNN and the iii-th vertex has the weight viv_ivi. The root of each tree is vertex 1. Given KKK key numbers x1,…,xkx_1,\dots,x_kx1,…,xk, find the number of solutions that remove exactly one number so that the weight of the lowest common ancestor of the vertices in A with the remaining key numbers is greater than the weight of the lowest common ancestor of the vertices in B with the remaining key numbers.
输入描述:
The first line has two positive integers N,K(2≤K≤N≤105)N,K (2 \leq K \leq N \leq 10^5)N,K(2≤K≤N≤105).The second line has KKK unique positive integers x1,…,xK(xi≤N)x_1,\dots,x_K (x_i \leq N)x1,…,xK(xi≤N).The third line has NNN positive integers ai(ai≤109)a_i (a_i \leq 10^9)ai(ai≤109) represents the weight of vertices in A.The fourth line has N−1N - 1N−1 positive integers {pai}\{pa_i\}{pai}, indicating that the number of the father of vertices i+1i+1i+1 in tree A is paipa_ipai.The fifth line has nnn positive integers bi(bi≤109)b_i (b_i \leq 10^9)bi(bi≤109) represents the weight of vertices in B.The sixth line has N−1N - 1N−1 positive integers {pbi}\{pb_i\}{pbi}, indicating that the number of the father of vertices i+1i+1i+1 in tree B is pbipb_ipbi.
输出描述:
One integer indicating the answer.
示例1
输入
5 3 5 4 3 6 6 3 4 6 1 2 2 4 7 4 5 7 7 1 1 3 2
输出
1
说明
In first case, the key numbers are 5,4,3. Remove key number 5, the lowest common ancestors of the vertices in A with the remaining key numbers is 2, in B is 3. Remove key number 4, the lowest common ancestors of the vertices in A with the remaining key numbers is 2, in B is 1. Remove key number 3, the lowest common ancestors of the vertices in A with the remaining key numbers is 4, in B is 1. Only remove key number 5 satisfies the requirement.
示例2
输入
10 3 10 9 8 8 9 9 2 7 9 0 0 7 4 1 1 2 4 3 4 2 4 7 7 7 2 3 4 5 6 1 5 3 1 1 3 1 2 4 7 3 5
输出
2
备注:
The lowest common ancestor (LCA) (also called least common ancestor) of two nodes v and w in a tree or directed acyclic graph (DAG) T is the lowest (i.e. deepest) node that has both v and w as descendants, where we define each node to be a descendant of itself (so if v has a direct connection from w, w is the lowest common ancestor).(From Wiki.)
思路：
1，lca的板子题稍微进阶点，参看lca的文章
2，用两棵树存各自的前缀后缀lca，然后调用即可
代码：
int n,k;
int vea[maxj],veb[maxj];
int lg[maxj];
int x[maxj];
void dd(){for(int i=1;i<=n;++i)lg[i]=lg[i-1]+(1<<lg[i-1]==i);
}
struct node{int pp[maxj],las[maxj];
int dep[maxj],vis[maxj],fa[maxj][100];
vector<int>g[maxj];
void add(int u,int v){g[u].emplace_back(v);
}
void dfs(int now ,int pre){dep[now]=dep[pre]+1;fa[now][0]=pre;for(int i=1;(1<<i)<=dep[now];++i){fa[now][i]=fa[fa[now][i-1]][i-1];}for(int i=0;i<g[now].size();++i){if(pre==g[now][i])continue;dfs(g[now][i],now);}
}int lca(int x,int y){if(dep[x]<dep[y])swap(x,y);while(dep[x]>dep[y])x=fa[x][lg[dep[x]-dep[y]]-1];if(x==y)return x;for(int k=lg[dep[x]];k>=0;--k){if(fa[x][k]!=fa[y][k]){x=fa[x][k];y=fa[y][k];}}return fa[x][0];
}
void dolca(){dfs(1,-1);pp[1]=x[1];for(int i=2;i<=k;++i){pp[i]=lca(pp[i-1],x[i]);}las[k]=x[k];for(int i=k-1;i>=1;--i){las[i]=lca(las[i+1],x[i]);}
}
int asklca(int w){if(w==1)return las[2];if(w==k)return pp[k-1];return lca(pp[w-1],las[w+1]);
}
}treea,treeb;//分treea和treeb两个对象void solve(){cin>>n>>k;//用struct封装一下，函数重用for(int i=1;i<=k;++i)cin>>x[i];for(int i=1;i<=n;++i)cin>>vea[i];for(int i=2;i<=n;++i){int pa;cin>>pa;treea.add(pa,i);}for(int i=1;i<=n;++i)cin>>veb[i];for(int i=2;i<=n;++i){int pa;cin>>pa;treeb.add(pa,i);}dd();treea.dolca();treeb.dolca();int ans=0;
// for(int i=1;i<=k;++i)cout<<treea.las[i]<<' ';for(int i=1;i<=k;++i){int aa=treea.asklca(i);int bb=treeb.asklca(i);
// cout<<aa<<' '<<bb<<'\n';if(vea[aa]>veb[bb])ans++;}cout<<ans<<'\n';
}
