hdu 4812 D Tree 树的点分治

来源：互联网发布：中国援助朝鲜知乎编辑：程序博客网时间：2024/06/05 14:08

题意 : 给你一棵树，每个点有权值，问是否有一条路径的乘积是K 。

思路 : 基于点的分治，对于一个重心，我们枚举以重心为根路径，计算出每个点从路径到根的乘积，然后对于这个乘积 Val，我们通过hash查找 K * inv[Val] 是否存在。hash[Val] 我存的到当前重心路径乘积为Val中下标最小的点，如果是INF则表示不存在，为了防止两个端点都在同一子树，要先访问完一遍子树，然后再更新hash。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#pragma comment(linker,"/STACK:102400000,102400000")typedef __int64 LL ;#define INF 0x3f3f3f3f#define MAXN 100005#define MOD 1000003int ans[2] ;int n , m , k ;bool vis[MAXN] ;int val[MAXN] ;int inv[MOD] ;int pi[MOD] ;struct Tree{    struct Edge{        int to , nex ;        Edge(){}        Edge( int _to , int _nex ) {            to = _to ;            nex = _nex ;        }    }edge[MAXN*2] ;    int head[MAXN] , Index ;    void init(){        memset( head , -1 , sizeof(head) ) ;        Index = 0 ;    }    void add( int from , int to ){        edge[Index] = Edge( to , head[from] ) ;        head[from] = Index ++ ;    }}tree;int sum[MAXN] , Max[MAXN] ;void dfs1( int u , int p ) {    sum[u] = 1 ;    Max[u] = 0 ;    for( int i = tree.head[u] ; ~i ; i = tree.edge[i].nex ) {        int ch = tree.edge[i].to ;        if( ch == p || vis[ch] ) continue ;        dfs1( ch , u  ) ;        sum[u] += sum[ch] ;        Max[u] = max( Max[u] , sum[ch] ) ;     }}int Min , Mini ;void dfs2( int u , int p , int ss ) {    int MM = max( Max[u] , ss - sum[u] ) ;    if( MM < Min ) {        Min = MM ; Mini = u ;    }    for( int i = tree.head[u] ; ~i ; i = tree.edge[i].nex ) {        int ch = tree.edge[i].to ;        if( ch == p || vis[ch] ) continue ;        dfs2( ch , u , ss ) ;    }}int getRoot( int u ) {    dfs1( u , -1 ) ;    Mini = -1 ;    Min = 999999999 ;    dfs2( u , -1 , sum[u] ) ;    return Mini ;}void dfs3( int u , int p , int Val ) {    Val = ( (LL)Val * val[u] ) % MOD ;    int key = ( (LL)k * inv[Val])%MOD ;    if( pi[ key ] != INF ) {        int tmp[2] ;        tmp[0] = u ;        tmp[1] = pi[ key ] ;        sort( tmp , tmp + 2 ) ;        if( (tmp[0] < ans[0]) || ( tmp[0] == ans[0] && tmp[1] < ans[1] )) {            ans[0] = tmp[0] ;            ans[1] = tmp[1] ;        }    }    for( int i = tree.head[u] ; ~i ; i = tree.edge[i].nex ) {        int ch = tree.edge[i].to ;        if( ch == p || vis[ch] ) continue ;        dfs3( ch , u , Val  ) ;    } }void dfs4( int u , int p , int Val ) {    Val = ( (LL)Val * val[u] ) % MOD ;    pi[Val] = min( pi[Val] , u ) ;    for( int i = tree.head[u] ; ~i ; i = tree.edge[i].nex ) {        int ch = tree.edge[i].to ;        if( ch == p || vis[ch] ) continue ;        dfs4( ch , u , Val  ) ;    }}void dfs5( int u , int p , int Val ) {    Val = ( (LL)Val * val[u] ) % MOD ;    pi[Val] = INF ;    for( int i = tree.head[u] ; ~i ; i = tree.edge[i].nex ) {        int ch = tree.edge[i].to ;        if( ch == p || vis[ch] ) continue ;        dfs5( ch , u , Val  ) ;    }}void solve( int u ) {    // 找重心     int root = getRoot( u ) ;    pi[val[root]] = root ;    for( int i = tree.head[root] ; ~ i ; i = tree.edge[i].nex ) {        // 计算各个顶点到根节点的π        int ch = tree.edge[i].to ;        if( vis[ch] ) continue ;        // 查询是否存在路径        dfs3( ch , root , 1 ) ;        // 在hash中添加路径        dfs4( ch , root , val[root] ) ;    }    for( int i = tree.head[root] ; ~i ; i = tree.edge[i].nex ) {        int ch = tree.edge[i].to ;        if( vis[ch] ) continue ;        dfs5( ch , root , val[root] ) ;    }    pi[val[root]] = INF ;    vis[root] = true ;    for( int i = tree.head[root] ; ~i ; i = tree.edge[i].nex ) {        int ch = tree.edge[i].to ;        if( vis[ch] ) continue ;        solve( ch ) ;    }}LL power( int a , int k ) {    if( k == 0 ) return 1 ;    else if( k == 1 ) return a ;    LL tmp = power( a , k >> 1 ) ;    if( k & 1 )         return ( ( tmp * tmp ) % MOD * a ) % MOD ;    return ( tmp * tmp ) % MOD ;}int main(){    for( int i = 0 ; i < MOD ; i ++ ) {        inv[i] = power( i , MOD - 2 ) ;    }    memset( pi  , INF , sizeof(pi) ) ;    while( scanf( "%d%d" , &n , &k ) != EOF ) {        tree.init() ;        for( int i = 1 ; i <= n ; i ++ ) {            scanf( "%d" , &val[i] ) ;        }        for( int i = 1 ; i < n ; i ++ ) {            int u , v ;            scanf( "%d%d" , &u , &v ) ;            tree.add( u , v ) ;            tree.add( v , u ) ;        }        ans[0] = INF ;        ans[1] = INF ;        memset( vis , false , sizeof(vis) ) ;        solve( 1 ) ;        if( ans[0] == INF ) {            puts( "No solution") ;        }else{            printf( "%d %d\n" , ans[0] , ans[1] ) ;        }    }    return 0 ;}

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