HYSBZ 1146 网络管理Network Treap+线段树+树链

来源：互联网发布：mysql 存储过程性能编辑：程序博客网时间：2024/06/05 19:57

题意很简单：在一颗树上修改某个点的值，然后查询两点之间路径上的第k大的元素

查阅了很多资料，因为不会Treap，链接如下：

http://blog.csdn.net/ssccode/article/details/17351461 (这个代码有点问题。比如未删除分配的空间。update顺序错误等。但是写成类的形式很爽。)
http://blog.csdn.net/acdreamers/article/details/11309971
http://www.nocow.cn/index.php/Treap

注：如果rotate看不懂的话你需要注意参数 node *&x的意义

代码有点长，如下:

//author: CHC//First Edit Time:  2015-06-06 09:14#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <set>#include <vector>#include <map>#include <queue>#include <set>#include <algorithm>#include <limits>using namespace std;typedef long long LL;const int MAXN=80000+1000;const int INF = numeric_limits<int>::max();const LL LL_INF= numeric_limits<LL>::max();class Treap {    public:        struct node {            node *ch[2];            int v,p,sz;            node(int _v,node *_n){ v=_v; ch[0]=ch[1]=_n; p=rand(); sz=1; }            void update(){sz=ch[0]->sz+ch[1]->sz+1;}            void fixch(node *left,node *right){ch[0]=left;ch[1]=right;}        };        node *root,*null;        Treap(){            null=new node(0,0);            null->fixch(null,null); null->sz=0;            root=null;        }        void rotate(node *&t,bool d){            node *_t=t->ch[d];            t->ch[d]=_t->ch[!d];            _t->ch[!d]=t;            t->update();            _t->update();            t=_t;        }        void __insert(node *&t,int val){            if(t==null){                t=new node(val,null);                return ;            }            bool d=(t->v < val);            __insert(t->ch[d],val);            if(t->ch[d]->p < t->p)rotate(t,d);            t->update();        }        void __del(node *&t,int val){            if(t==null)return ;            if(t->v==val){                bool d=t->ch[1]->p < t->ch[0]->p;                if(t->ch[d]==null){                    node *tmp=t;                    t=t->ch[!d];                    delete tmp;                    return ;                }                rotate(t,d);                __del(t->ch[!d],val);            }            else {                bool d=(t->v < val);                __del(t->ch[d],val);            }            t->update();        }        int __rank(node *t,int val){            if(t==null)return 0;            if(t->v < val)return __rank(t->ch[1],val);            return t->ch[1]->sz+1+__rank(t->ch[0],val);        }        void __clean(node *&t){            if(t==null)return ;            if(t->ch[0]!=null)__clean(t->ch[0]);            if(t->ch[1]!=null)__clean(t->ch[1]);            delete t;            t=null;        }        void insert(int val){ __insert(root,val); }        void del(int val){ __del(root,val); }        int rank(int val){ return __rank(root,val);}        void clean(){ __clean(root); }        ~Treap(){ clean(); delete null; }};struct Edge {    int to,next;    Edge(){}    Edge(int _to,int _next):to(_to),next(_next){}}e[MAXN<<1];int head[MAXN],tot;void AddEdge(int u,int v){    e[tot]=Edge(v,head[u]);    head[u]=tot++;    e[tot]=Edge(u,head[v]);    head[v]=tot++;}int dep[MAXN],fa[MAXN],son[MAXN],sz[MAXN];int top[MAXN],tid[MAXN],ranked[MAXN],tim;void init(){    memset(head,-1,sizeof(head));    tim=tot=0;}void dfs1(int u,int father,int d){    fa[u]=father;dep[u]=d;son[u]=-1;sz[u]=1;    for(int i=head[u];~i;i=e[i].next){        int v=e[i].to;        if(fa[u]!=v){            dfs1(v,u,d+1);            sz[u]+=sz[v];            if(son[u]==-1||sz[v]>sz[son[u]])son[u]=v;        }    }}void dfs2(int u,int tp){    top[u]=tp;tid[u]=++tim;ranked[tid[u]]=u;    if(son[u]==-1)return ;    dfs2(son[u],tp);    for(int i=head[u];~i;i=e[i].next){        int v=e[i].to;        if(v!=son[u]&&v!=fa[u]) dfs2(v,v);    }}Treap treap[MAXN<<2];int A[MAXN];#define lson L,mid,rt<<1#define rson mid+1,R,rt<<1|1void build(int L,int R,int rt){    treap[rt].clean();    for(int i=L;i<=R;i++){        treap[rt].insert(A[ranked[i]]);    }    if(L==R)return ;    int mid=(L+R)>>1;    build(lson);build(rson);}int query(int L,int R,int rt,int l,int r,int v){    if(l<=L&&R<=r){        return treap[rt].rank(v);    }    int mid=(L+R)>>1,ans=0;    if(l<=mid)ans+=query(lson,l,r,v);    if(r>mid)ans+=query(rson,l,r,v);    return ans;}void update(int L,int R,int rt,int pos,int pre,int now){    treap[rt].del(pre);    treap[rt].insert(now);    if(L==R)return ;    int mid=(L+R)>>1;    if(pos<=mid)update(lson,pos,pre,now);    else update(rson,pos,pre,now);}int n,q;int queryx(int x,int y,int v){    int ans=0;    while(top[x]!=top[y]){        if(dep[top[x]]<dep[top[y]])swap(x,y);        ans+=query(1,n,1,tid[top[x]],tid[x],v);        x=fa[top[x]];    }    if(dep[x]>dep[y])swap(x,y);    ans+=query(1,n,1,tid[x],tid[y],v);    return ans;}void updatex(int pos,int v){    update(1,n,1,tid[pos],A[pos],v);    A[pos]=v;}int main(){    while(~scanf("%d%d",&n,&q)){        init();        for(int i=1;i<=n;i++)scanf("%d",&A[i]);        for(int i=0,x,y;i<n-1;i++){            scanf("%d%d",&x,&y);            AddEdge(x,y);        }        dfs1(1,0,0);        //puts("here");        dfs2(1,1);        build(1,n,1);        for(int i=0,x,y,k;i<q;i++){            scanf("%d%d%d",&k,&x,&y);            if(k){                int l=0,r=1e+9,ans=-1;                while(l<=r){                    int mid=(l+r)>>1;                    int tmp1=queryx(x,y,mid);                    if(tmp1<k){                        ans=mid;                        r=mid-1;                    }                    else l=mid+1;                }                if(ans-1>=0) printf("%d\n",ans-1);                else puts("invalid request!");            }            else updatex(x,y);        }    }    return 0;}

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