HDU4670 Cube number on a tree

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4670

算法：基于点的树链分治

思路：

//树基于点的分治算法，可以参见国家集训队论文：2009年漆子超《分治算法在树的路径问题中的应用》
/*
000111222
+ 012012012
----------------------
= 012120201

*/

//Mul[u,v] 等于点u到点v的乘积中，各个素数的个数对3取余后的一个数列

//对于每一个Mul[u, S], 求Mul(S, v]+Mul[u,S]=0的个数(S为分治点)

//用map记录下所有Mul[S,u]的状态（用longlong保存3进制数），再用一个map保存取出分治点S以后(S,v]的状态，对于第一个map中的每一个状态，求在第二个map中是否存在它的互补状态

注意：

//第一次遇到手动扩栈的题，检查了好长时间
//素数有可能超过int

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<cstdio>#include<cstring>#include<map>using namespace std;#define LL __int64const int M = 50010;LL pri[35];//保存素数int Num;//素数的个数int num[M][30];//每个数的化简struct Edge {    int v, next;} edge[M << 1];int head[M], E;//邻接表参数LL ans = 0;//保存最终结果int mmax, root, size[M], mx[M], vis[M];//求重点的参数int tmp_cnt;//计算类似8,27,125这样单独成区间的区间个数int dis[M][30];//保存区间[S,i]中第j个素数的个数dis[i][j]map<LL, LL> er, mp;//er是计算3进制下需要匹配的数的个数，mp是计算分治点S到目标点Q,Mul[S,Q]在3进制下得到的数map<LL, LL>::iterator it;void init() {    ans = E = 0;    memset(vis, 0, sizeof(vis));    memset(head, -1, sizeof(head));    memset(num, 0, sizeof(num));}void add_edge(int s, int v) {    edge[E].v = v;    edge[E].next = head[s];    head[s] = E++;}void dfs_uu(int u, int fa) {    size[u] = 1;    mx[u] = 0;    for (int i = head[u]; i != -1; i = edge[i].next) {        int v = edge[i].v;        if (v == fa || vis[v])continue;        dfs_uu(v, u);        size[u] += size[v];        if (size[v] > mx[u]) mx[u] = size[v];    }}void dfs_u(int u, int fa, int r) {    if (size[r] - size[u] > mx[u]) mx[u] = size[r] - size[u];    if (mmax > mx[u]) { mmax = mx[u]; root = u;}    for (int i = head[u]; i != -1; i = edge[i].next) {        int v = edge[i].v;        if (v == fa || vis[v]) continue;        dfs_u(v, u, r);    }}//以上两个函数求重点void dfs_dis(int u, int fa, int r) {    LL tmp = 0, sum = 0, sum_nee = 0;    for (int i = 0; i < Num; i++) {        dis[u][i] = dis[fa][i] + num[u][i];        if (dis[u][i] > 2) dis[u][i] -= 3;        tmp = 3 - dis[u][i] + dis[r][i]; if (tmp >= 3) tmp -= 3;        sum = sum * 3 + tmp;        sum_nee = sum_nee * 3 + dis[u][i];    }    if (sum == sum_nee) tmp_cnt++;    er[sum]++;    mp[sum_nee]++;    for (int i = head[u]; i != -1; i = edge[i].next) {        int v = edge[i].v;        if (v == fa || vis[v])            continue;        dfs_dis(v, u, r);    }}LL cala(int u, int fa) {    mp.clear();    er.clear();    LL tt = 0;    tmp_cnt = 0;    if (fa != 0) for (int i = 0; i < Num; i++) dis[fa][i] = num[fa][i];    dfs_dis(u, fa, fa == 0 ? u : fa);    for (it = er.begin(); it != er.end(); it++) {        if (mp.find(it->first) != mp.end())            tt += it->second * mp[it->first];    }    return (tt - tmp_cnt) / 2 + tmp_cnt;}//以上两个【核心】函数求分治区间中满足条件的点对~void solve(int u) {    int rt = 0;    mmax = 123456;    root = u;    dfs_uu(u, -1);    dfs_u(u, -1, u);    ans += cala(root, 0);    vis[root] = 1;    rt = root;    for (int i = head[rt]; i != -1; i = edge[i].next) {        int v = edge[i].v;        if (vis[v]) continue;        ans -= cala(v, rt);        solve(v);    }}int main() {    int n;    while (scanf("%d", &n) != EOF) {        int i, j;        LL val;        init();        scanf("%d", &Num);        for (i = 0; i < Num; i++) {            scanf("%I64d", &pri[i]);        }        for (i = 1; i <= n; i++) {            scanf("%I64d", &val);            for (j = 0; j < Num; j++) {                while (val % pri[j] == 0) {                    num[i][j]++;                    val = val / pri[j];                    if (num[i][j] > 2)                        num[i][j] -= 3;                }            }        }        int a, b;        for (i = 0; i < n - 1; i++) {            scanf("%d%d", &a, &b);            add_edge(a, b);            add_edge(b, a);        }        solve(1);        printf("%I64d\n", ans);    }    return 0;}

	
				
	
					   
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