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 58 5
105 2 9 3 8 5 7 7
1 2
1 3
1 4
3 5
3 6
3 7
4 8
2 5 1
2 5 2
2 5 3
2 5 4
7 8 2 Output:
2
8
9
105
7

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)))

[cpp]
#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]];

补充：软件开发 , C++ ,