SPOJ 10628(COT-树上第k大-函数式线段树)

SPOJ Problem Set (classical)

10628. Count on a tree

Problem code: COT

You are given a tree with N nodes.The tree nodes are numbered from 1 to N.Each node has an integer weight.

We will ask you to perform the following operation:

• u v k : ask for the kth minimum weight on the path from node u to node v

 

Input

In the first line there are two integers N and M.(N,M<=100000)

In the second line there are N integers.The ith integer denotes the weight of the ith node.

In the next N-1 lines,each line contains two integers u v,which describes an edge (u,v).

In the next M lines,each line contains three integers u v k,which means an operation asking for the kth minimum weight on the path from node u to node v.

Output

For each operation,print its result.

Example

Input:
8 5
8 5105 2 9 3 8 5 7 71 2        1 31 43 53 63 74 82 5 12 5 22 5 32 5 47 8 2 

Output:2891057 

我们在树上每个结点塞一个函数式线段树。

表示从这个节点到根的链所代表的权值线段树

则(u-v)=(u)+(v)-(lca(u,v))-(fa(lca(u,v)))

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#include<cmath>#include<cctype>#include<map>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForD(i,n) for(int i=n;i;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define RepD(i,n) for(int i=n;i>=0;i--)#define MEM(a) memset(a,0,sizeof(a))#define MEMI(a) memset(a,127,sizeof(a))#define MEMi(a) memset(a,128,sizeof(a))#define MAXN (100000+10)#define MAXM (200000+10)#define Li (18)int n,m,w[MAXN];int edge[MAXM],next[MAXM]={0},pre[MAXN]={0},size=0;void addedge(int u,int v){edge[++size]=v;next[size]=pre[u];pre[u]=size;}void addedge2(int u,int v){addedge(u,v),addedge(v,u);}struct node{int ch[2],c;node(){ch[0]=ch[1]=c=0;}}q[MAXN*Li];int root[MAXN],tail=0;void ins(int &x,int p,int l,int r,int c){x=++tail;if (p) q[x]=q[p];q[x].c++;if (l==r) return;int m=(l+r)>>1;if (c<=m) ins(q[x].ch[0],q[x].ch[0],l,m,c);else ins(q[x].ch[1],q[x].ch[1],m+1,r,c);}map<int ,int> h;int a2[MAXN],a2_size;int d[MAXN],father[MAXN][Li]={0};int q2[MAXN];void bfs(){int head=1,tail=1;q2[1]=1;d[1]=1;ins(root[1],root[1],1,a2_size,w[1]);while (head<=tail){int x=q2[head++];For(i,Li-1) {father[x][i]=father[father[x][i-1]][i-1];if (!father[x][i]) break;}Forp(x){int &v=edge[p];if (d[v]) continue;d[v]=d[x]+1;father[v][0]=x;q2[++tail]=v;ins(root[v],root[x],1,a2_size,w[v]);}}}int bin[MAXN];int lca(int u,int v){if (d[u]<d[v]) swap(u,v);int t=d[u]-d[v];Rep(i,Li)if (bin[i]&t) u=father[u][i];int i=Li-1;while (u^v){while (i&&father[u][i]==father[v][i]) i--;u=father[u][i],v=father[v][i];}return u;}int ans[MAXN],ans_siz=0,ans_end=0;int getsum(){int tot=0;For(i,ans_end) tot+=q[q[ans[i]].ch[0]].c;Fork(i,ans_end+1,ans_siz) tot-=q[q[ans[i]].ch[0]].c;return tot;}void turn(bool c){For(i,ans_siz) ans[i]=q[ans[i]].ch[c];}int main(){//freopen("spoj10628_COT.in","r",stdin);//freopen(".out","w",stdout);bin[0]=1;For(i,Li-1) bin[i]=bin[i-1]<<1;scanf("%d%d",&n,&m);For(i,n) scanf("%d",&w[i]),a2[i]=w[i];sort(a2+1,a2+1+n);int size=unique(a2+1,a2+1+n)-(a2+1);a2_size=size;For(i,size) h[a2[i]]=i;For(i,n) w[i]=h[w[i]];For(i,n-1) {int u,v;scanf("%d%d",&u,&v);addedge2(u,v);}bfs();For(i,m){int u,v,k;scanf("%d%d%d",&u,&v,&k);int f=lca(u,v);ans[1]=root[u],ans[2]=root[v],ans[3]=root[f],ans[4]=root[father[f][0]];ans_siz=4,ans_end=2;int l=1,r=size;while (l<r){int m=l+r>>1,s;if (k<=(s=getsum())) r=m,turn(0);else l=m+1,k-=s,turn(1); }printf("%d\n",a2[l]);}return 0;}