SPOJ Count on a tree

Count on a tree

Time Limit:129MS Memory Limit:1572864KB 64bit IO Format:%lld & %llu

Submit Status Practice SPOJ COT

Description

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 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 
分析：tarjan（st表）+主席树，加了一下读入挂，挺快的；
代码：
st表：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#include <set>

#include <map>

#include <queue>

#include <stack>

#include <vector>

#include <list>

#define rep(i,m,n) for(i=m;i<=n;i++)

#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)

#define mod 1000000007

#define inf 0x3f3f3f3f

#define vi vector<int>

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define ll long long

#define pi acos(-1.0)

#define pii pair<int,int>

#define Lson L, mid, rt<<1

#define Rson mid+1, R, rt<<1|1

const int maxn=4e6+;

using namespace std;

ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}

ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}

int n,m,k,t,a[maxn],b[maxn],s[maxn],ls[maxn],rs[maxn],root[maxn],h[maxn],p[maxn],dep[maxn],vis[maxn],sz,num,tot,all;

pii st[][maxn];

inline ll read()

{

    ll x=;int f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

void init()

{

    for(int i=;i<=all;i++)p[i]=+p[i>>];

    for(int i=;i<=;i++)

        for(int j=;(ll)j+(<<i)-<=all;j++)

            st[i][j]=min(st[i-][j],st[i-][j+(<<(i-))]);

}

int query(int l,int r)

{

    int x=p[r-l+];

    return min(st[x][l],st[x][r-(<<x)+]).se;

}

struct node

{

    int to,nxt;

}e[maxn];

void add(int x,int y)

{

    tot++;

    e[tot].to=y;

    e[tot].nxt=h[x];

    h[x]=tot;

}

void insert(int l,int r,int x,int &y,int v)

{

    y=++sz;

    s[y]=s[x]+;

    if(l==r)return;

    ls[y]=ls[x],rs[y]=rs[x];

    int mid=l+r>>;

    if(v<=mid)insert(l,mid,ls[x],ls[y],v);

    else insert(mid+,r,rs[x],rs[y],v);

}

int gao(int l,int r,int x,int y,int z,int k,int care)

{

    if(l==r)return l;

    int mid=l+r>>,j;

    j=s[ls[y]]+s[ls[z]]-*s[ls[x]]+(care>=l&&care<=mid);

    if(j>=k)return gao(l,mid,ls[x],ls[y],ls[z],k,care);

    else return gao(mid+,r,rs[x],rs[y],rs[z],k-j,care);

}

void dfs(int now,int pre)

{

    st[][++all]=mp(dep[now],now);

    vis[now]=all;

    insert(,num,root[pre],root[now],a[now]);

    for(int i=h[now];i;i=e[i].nxt)

    {

        int to=e[i].to;

        if(to!=pre)

        {

            dep[to]=dep[now]+;

            dfs(to,now);

            st[][++all]=mp(dep[now],now);

        }

    }

}

int main()

{

    int i,j;

    scanf("%d%d",&n,&m);

    rep(i,,n)a[i]=read(),b[i]=a[i];

    sort(b+,b+n+);

    num=unique(b+,b+n+)-b-;

    rep(i,,n)a[i]=lower_bound(b+,b+num+,a[i])-b;

    rep(i,,n-)

    {

        int c,d;

        c=read(),d=read();

        add(c,d);add(d,c);

    }

    dfs(,);

    init();

    rep(i,,m)

    {

        int c,d,k;

        c=read(),d=read(),k=read();

        if(vis[c]>vis[d])swap(c,d);

        int fa=query(vis[c],vis[d]);

        printf("%d\n",b[gao(,num,root[fa],root[c],root[d],k,a[fa])]);

    }

    //system("Pause");

    return ;

}

Tarjan：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#include <set>

#include <map>

#include <queue>

#include <stack>

#include <vector>

#include <list>

#define rep(i,m,n) for(i=m;i<=n;i++)

#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)

#define mod 1000000007

#define inf 0x3f3f3f3f

#define vi vector<int>

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define ll long long

#define pi acos(-1.0)

#define pii pair<int,int>

#define Lson L, mid, rt<<1

#define Rson mid+1, R, rt<<1|1

const int maxn=4e6+;

using namespace std;

ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}

ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}

int n,m,k,t,a[maxn],b[maxn],s[maxn],ls[maxn],rs[maxn],root[maxn],vis[maxn],ans[maxn],fa[maxn],h[maxn],g[maxn],sz,num,tot,tot1;

struct node

{

    int x,y,z;

    node(){}

    node(int _x,int _y,int _z):x(_x),y(_y),z(_z){};

};

struct node1

{

    int to,nxt;

}e[maxn];

struct node2

{

    node p;

    int nxt;

}f[maxn];

void add(int x,int y)

{

    tot++;

    e[tot].to=y;

    e[tot].nxt=h[x];

    h[x]=tot;

}

void add1(int x,int y,int z,int k)

{

    tot1++;

    f[tot1].p=node(y,z,k);

    f[tot1].nxt=g[x];

    g[x]=tot1;

}

inline ll read()

{

    ll x=;int f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int find(int x)

{

    return fa[x]==x?x:fa[x]=find(fa[x]);

}

void insert(int l,int r,int x,int &y,int v)

{

    y=++sz;

    s[y]=s[x]+;

    if(l==r)return;

    ls[y]=ls[x],rs[y]=rs[x];

    int mid=l+r>>;

    if(v<=mid)insert(l,mid,ls[x],ls[y],v);

    else insert(mid+,r,rs[x],rs[y],v);

}

int gao(int l,int r,int x,int y,int z,int k,int care)

{

    if(l==r)return l;

    int mid=l+r>>,j;

    j=s[ls[y]]+s[ls[z]]-*s[ls[x]]+(care>=l&&care<=mid);

    if(j>=k)return gao(l,mid,ls[x],ls[y],ls[z],k,care);

    else return gao(mid+,r,rs[x],rs[y],rs[z],k-j,care);

}

void dfs(int now,int pre)

{

    vis[now]=;

    insert(,num,root[pre],root[now],a[now]);

    for(int i=g[now];i;i=f[i].nxt)

    {

        node p=f[i].p;

        if(vis[p.y])

        {

            int fa=find(p.y);

            ans[p.x]=b[gao(,num,root[fa],root[now],root[p.y],p.z,a[fa])];

        }

    }

    for(int i=h[now];i;i=e[i].nxt)

    {

        int to=e[i].to;

        if(!vis[to])

        {

            dfs(to,now);

            fa[to]=now;

        }

    }

}

int main()

{

    int i,j;

    scanf("%d%d",&n,&m);

    rep(i,,n)a[i]=read(),b[i]=a[i],fa[i]=i;

    sort(b+,b+n+);

    num=unique(b+,b+n+)-b-;

    rep(i,,n)a[i]=lower_bound(b+,b+num+,a[i])-b;

    rep(i,,n-)

    {

        int c,d;

        c=read(),d=read();

        add(c,d);add(d,c);

    }

    rep(i,,m)

    {

        int c,d,k;

        c=read(),d=read(),k=read();

        add1(c,i,d,k);

        add1(d,i,c,k);

    }

    dfs(,);

    rep(i,,m)printf("%d\n",ans[i]);

    //system("Pause");

    return ;

}

巴特西

SPOJ Count on a tree

Input

Output

Example

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