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  • 【BZOJ1803】Spoj1487 Query on a tree III 主席树+DFS序

    【BZOJ1803】Spoj1487 Query on a tree III

    Description

    You are given a node-labeled rooted tree with n nodes. Define the query (x, k): Find the node whose label is k-th largest in the subtree of the node x. Assume no two nodes have the same labels.

    Input

    The first line contains one integer n (1 <= n <= 10^5). The next line contains n integers li (0 <= li <= 109) which denotes the label of the i-th node. Each line of the following n - 1 lines contains two integers u, v. They denote there is an edge between node u and node v. Node 1 is the root of the tree. The next line contains one integer m (1 <= m <= 10^4) which denotes the number of the queries. Each line of the next m contains two integers x, k. (k <= the total node number in the subtree of x)

    Output

    For each query (x, k), output the index of the node whose label is the k-th largest in the subtree of the node x.

    Sample Input

    5
    1 3 5 2 7
    1 2
    2 3
    1 4
    3 5
    4
    2
    3
    4 1
    3 2
    3 2

    Sample Output

    5 4 5 5

    题解:这样例你告诉我是求子树第k大?(the k_th largest)分明是第k小啊!

    直接主席树+DFS序搞定。

    #include <cstdio> 
    #include <cstring> 
    #include <iostream> 
    #include <algorithm> 
    using namespace std; 
    const int maxn=100010; 
    int n,m,cnt,tot; 
    int to[maxn<<1],next[maxn<<1],head[maxn],v[maxn],p[maxn],q[maxn],rt[maxn]; 
    int ls[maxn*30],rs[maxn*30],siz[maxn*30],s[maxn*30]; 
    struct number 
    { 
        int val,org; 
    }num[maxn]; 
    bool cmp(number a,number b) 
    { 
        return a.val<b.val; 
    } 
    void insert(int x,int &y,int l,int r,int a,int b) 
    { 
        if(l>r)  return ; 
        y=++tot,siz[y]=siz[x]+1; 
        if(l==r) 
        { 
            s[y]=b; 
            return ; 
        } 
        int mid=l+r>>1; 
        if(a<=mid)   rs[y]=rs[x],insert(ls[x],ls[y],l,mid,a,b); 
        else    ls[y]=ls[x],insert(rs[x],rs[y],mid+1,r,a,b); 
    } 
    int query(int a,int b,int l,int r,int c) 
    { 
        if(l==r)    return s[a]; 
        int mid=l+r>>1,sm=siz[ls[a]]-siz[ls[b]]; 
        if(c<=sm)    return query(ls[a],ls[b],l,mid,c); 
        else    return query(rs[a],rs[b],mid+1,r,c-sm); 
    } 
    void dfs(int x,int fa) 
    { 
        p[x]=++p[0]; 
        insert(rt[p[0]-1],rt[p[0]],1,n,v[x],x); 
        for(int i=head[x];i!=-1;i=next[i])  if(to[i]!=fa)   dfs(to[i],x); 
        q[x]=p[0]; 
    } 
    void add(int a,int b) 
    { 
        to[cnt]=b,next[cnt]=head[a],head[a]=cnt++; 
    } 
    int main() 
    { 
        scanf("%d",&n); 
        int i,a,b; 
        for(i=1;i<=n;i++)    scanf("%d",&num[i].val),num[i].org=i; 
        sort(num+1,num+n+1,cmp); 
        for(i=1;i<=n;i++)    v[num[i].org]=i; 
        memset(head,-1,sizeof(head)); 
        for(i=1;i<n;i++) 
        { 
            scanf("%d%d",&a,&b); 
            add(a,b),add(b,a); 
        } 
        dfs(1,0); 
        scanf("%d",&m); 
        for(i=1;i<=m;i++) 
        { 
            scanf("%d%d",&a,&b); 
            printf("%d
    ",query(rt[q[a]],rt[p[a]-1],1,n,b)); 
        } 
        return 0; 
    }
    
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  • 原文地址:https://www.cnblogs.com/CQzhangyu/p/6952518.html
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