【codeforces 983E】NN country
Description
In the NN country, there are n cities, numbered from 1 to n, and n−1 roads, connecting them. There is a roads path between any two cities.
There are m bidirectional bus routes between cities. Buses drive between two cities taking the shortest path with stops in every city they drive through. Travelling by bus, you can travel from any stop on the route to any other. You can travel between cities only by bus.
You are interested in q questions: is it possible to get from one city to another and what is the minimum number of buses you need to use for it?
Input
The first line contains a single integer n (2≤n≤2⋅105) — the number of cities.
The second line contains n−1 integers p2,p3,…,pn (1≤pi<i), where pi means that cities pi and i are connected by road.
The third line contains a single integer m (1≤m≤2⋅105) — the number of bus routes.
Each of the next m m lines contains 2 integers a and b (1≤a,b≤n, a≠b), meaning that there is a bus route between cities a and b. It is possible that there is more than one route between two cities.
The next line contains a single integer q (1≤q≤2⋅105) — the number of questions you are interested in.
Output
Print the answer for each question on a separate line. If there is no way to get from one city to another, print −1. Otherwise print the minimum number of buses you have to use.
题意：给定一棵 $n$ 个点的树， $m$ 条链， $q$ 个询问，每次询问 $a$ 到 $b$ 之间的路径最少可用几条给定链完全覆盖，无解输出 $-1$ 。
分析：
对于每一条 $a$ 与$b$ 间的路径，都可拆分为 $a$ 到 $lca$ 的路径和 $b$ 到 $lca$ 的路径
采用贪心策略， $low[x][i]$ 表示从 $x$ 点出发向上选择不超过 $2^i$ 条链可抵达的深度最浅的点。这时对于每一个询问可将询问的两个端点修改为利用贪心策略跳到的深度大于 $lca$ 且深度最小的节点，并记录下答案，这个过程可以用倍增完成。注意特判端点即 $lca$ 的情况。
然后出现两种情况。若修改后的两个端点出现在同一条给定链上，答案为原答案 $+1$ ，否则答案为原答案 $+2$ 。问题模型转换为，每次询问一个点对是否出现在同一条给定链上。记录下 $dfs$ 序，在深搜过程中利用树状数组统计即可。
时间复杂度 $O(nlogn)$ 。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<vector> 5 #define LL long long 6 using namespace std; 7 const int N=2e5+5; 8 const int inf=0x3f3f3f3f; 9 int n,m,Q,cnt,val,x,y,ind; 10 int deep[N],in[N],out[N],last[N]; 11 int first[N],ans[N],tr[N]; 12 int fa[N][20],low[N][20]; 13 bool ok[N]; 14 vector<int> a[N],b[N]; 15 struct edge{int to,next;}e[N]; 16 struct chain{int x,y,t;}c[N],q[N]; 17 int read() 18 { 19 int x=0,f=1;char c=getchar(); 20 while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} 21 while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} 22 return x*f; 23 } 24 void ins(int u,int v){e[++cnt]=(edge){v,first[u]};first[u]=cnt;} 25 void dfs(int x) 26 { 27 in[x]=++ind; 28 for(int i=1;(1<<i)<=deep[x];i++) 29 fa[x][i]=fa[fa[x][i-1]][i-1]; 30 for(int i=first[x];i;i=e[i].next) 31 { 32 deep[e[i].to]=deep[x]+1; 33 dfs(e[i].to); 34 } 35 out[x]=ind; 36 } 37 int lca(int x,int y) 38 { 39 if(deep[x]<deep[y])swap(x,y); 40 int d=deep[x]-deep[y]; 41 for(int i=0;(1<<i)<=d;i++) 42 if((1<<i)&d)x=fa[x][i]; 43 if(x==y)return x; 44 for(int i=17;i>=0;i--) 45 if((1<<i)<=deep[x]&&fa[x][i]!=fa[y][i]) 46 x=fa[x][i],y=fa[y][i]; 47 return fa[x][0]; 48 } 49 void dfslow(int x) 50 { 51 for(int i=first[x];i;i=e[i].next) 52 { 53 int to=e[i].to;dfslow(to); 54 if(deep[low[to][0]]<deep[low[x][0]]) 55 low[x][0]=low[to][0]; 56 } 57 } 58 int find(int x,int t) 59 { 60 if(deep[low[x][17]]>deep[t]){val=-inf;return -1;} 61 if(x==t){val=-1;return 0;} 62 val=0; 63 for(int i=17;i>=0;i--) 64 if(deep[low[x][i]]>deep[t]) 65 x=low[x][i],val|=(1<<i); 66 return x; 67 } 68 int lowbit(int x){return x&(-x);} 69 void add(int x,int v){for(;x<=n;x+=lowbit(x))tr[x]+=v;} 70 int query(int x){int ans=0;for(;x;x-=lowbit(x))ans+=tr[x];return ans;} 71 void work(int x) 72 { 73 for(int sz=b[x].size(),i=0;i<sz;i++) 74 { 75 int t=b[x][i]; 76 last[t]=query(out[q[t].y])-query(in[q[t].y]-1); 77 } 78 for(int sz=a[x].size(),i=0;i<sz;i++)add(in[a[x][i]],1); 79 for(int i=first[x];i;i=e[i].next)work(e[i].to); 80 for(int sz=b[x].size(),i=0;i<sz;i++) 81 { 82 int t=b[x][i]; 83 if(query(out[q[t].y])-query(in[q[t].y]-1)!=last[t])ok[t]=true; 84 } 85 } 86 int main() 87 { 88 n=read(); 89 for(int i=2;i<=n;i++) 90 fa[i][0]=read(),ins(fa[i][0],i); 91 dfs(1); 92 for(int i=1;i<=n;i++)low[i][0]=i; 93 m=read(); 94 for(int i=1;i<=m;i++) 95 { 96 c[i].x=read();c[i].y=read(); 97 c[i].t=lca(c[i].x,c[i].y); 98 if(deep[c[i].t]<deep[low[c[i].x][0]]) 99 low[c[i].x][0]=c[i].t; 100 if(deep[c[i].t]<deep[low[c[i].y][0]]) 101 low[c[i].y][0]=c[i].t; 102 a[c[i].x].push_back(c[i].y); 103 a[c[i].y].push_back(c[i].x); 104 } 105 dfslow(1); 106 for(int t=1;t<=n;t++) 107 for(int i=1;i<=17;i++) 108 low[t][i]=low[low[t][i-1]][i-1]; 109 Q=read(); 110 for(int i=1;i<=Q;i++) 111 { 112 q[i].x=read();q[i].y=read(); 113 q[i].t=lca(q[i].x,q[i].y); 114 ans[i]=2; 115 x=find(q[i].x,q[i].t);ans[i]+=val; 116 y=find(q[i].y,q[i].t);ans[i]+=val; 117 if(x>0&&y>0) 118 { 119 q[i].x=x;q[i].y=y; 120 b[x].push_back(i); 121 } 122 } 123 work(1); 124 for(int i=1;i<=Q;i++) 125 if(ok[i])ans[i]--; 126 for(int i=1;i<=Q;i++) 127 printf("%d\n",ans[i]<0?-1:ans[i]); 128 return 0; 129 }
