HDU 4812 D Tree (树分治之点分治)

题目大意：

给出一棵树, 顶点数 n <= 100000

每个点有一个权值

给出K

询问这棵树中两点路径上的点的权值乘积对1e6 + 3取模之后等于K的路径的两个端点形成的点对中字典序最小的

大致思路：

点分治第二题

用f[i]表示从当前的子树根开始到某个点的路径乘积为i的点的最小标号

于是对于每一次分治, 去掉重心x之后, 依次处理每一棵子树来使得f的来源不一样

注意要预处理逆元

代码如下：

Result  :  Accepted     Memory  :  24420 KB     Time :  2823 ms

/* * Author: Gatevin * Created Time:  2015/10/14 9:59:26 * File Name: Sakura_Chiyo.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define maxn 100010struct Edge{    int u, v, nex;    Edge(){}    Edge(int _u, int _v, int _nex)    {        u = _u, v = _v, nex = _nex;    }};Edge edge[maxn << 1];int head[maxn];int E;int n;void add_Edge(int u, int v){    edge[++E] = Edge(u, v, head[u]);    head[u] = E;}int del[maxn];int root;int mx[maxn];int size[maxn];int mi;int N;lint K;lint w[maxn];const lint mod = 1e6 + 3;pair<int, int> ans;lint rev[1000010];void getRev(){    rev[1] = 1;    for(lint i = 2; i < mod; i++)        rev[i] = (mod - mod / i) * rev[mod % i] % mod;}void dfs_size(int now, int father){    size[now] = 1;    mx[now] = 1;    for(int i = head[now]; i + 1; i = edge[i].nex)    {        int v = edge[i].v;        if(v != father && !del[v])        {            dfs_size(v, now);            size[now] += size[v];            if(size[v] > mx[now]) mx[now] = size[v];        }    }}void dfs_root(int r, int now, int father){    if(size[r] - size[now] > mx[now]) mx[now] = size[r] - size[now];    if(mx[now] < mi) mi = mx[now], root = now;    for(int i = head[now]; i + 1; i = edge[i].nex)    {        int v = edge[i].v;        if(v != father && !del[v]) dfs_root(r, v, now);    }}lint f[1000010];//f[i]表示距离为i的字典序最小的点的标号+preconst pair<int, int> inf = make_pair(1e9, 1e9);lint pre;const lint bit = 1e5;void get(int now, int father, lint dis, int flag){    dis = dis*w[now] % mod;    if(flag == 1)    {        //dis*x % mod = K -> x = K*dis^(mod - 2) % mod        if(f[K*rev[dis] % mod] > pre)        {            int other = (int)(f[K*rev[dis] % mod] - pre);            pair<int, int> p = other < now ? make_pair(other, now) : make_pair(now, other);            if(ans > p) ans = p;        }    }    else    {        if(f[dis] <= pre) f[dis] = pre + now;        else f[dis] = min(f[dis], pre + now);    }    for(int i = head[now]; i + 1; i = edge[i].nex)    {        int v = edge[i].v;        if(!del[v] && v != father)            get(v, now, dis, flag);    }}void dfs(int now){    mi = N;    dfs_size(now, 0);    dfs_root(now, now, 0);    del[root] = 1;    pre += bit;    f[w[root]] = pre + root;//pre是因为每次清空f数组是不现实的, 于是用pre/bit表示第几次    for(int i = head[root]; i + 1; i = edge[i].nex)    {        int v = edge[i].v;        if(!del[v])        {            get(v, root, 1, 1);            get(v, root, w[root], 0);        }    }    for(int i = head[root]; i + 1; i = edge[i].nex)    {        int v = edge[i].v;        if(!del[v]) dfs(v);    }}void solve(){    ans = inf;    //pre = 0    //memset(f, 0, sizeof(f));利用pre不清0, f就可以不清0了    dfs(1);    if(ans == inf) puts("No solution");    else printf("%d %d\n", ans.first, ans.second);}int main(){    getRev();//预处理逆元    pre = 0;    while(scanf("%d %I64d", &N, &K) != EOF)    {        E = 0;        memset(head, -1, sizeof(head));        memset(del, 0, sizeof(del));                for(int i = 1; i <= N; i++)            scanf("%I64d", &w[i]);        int u, v;        for(int i = 1; i < N; i++)        {            scanf("%d %d", &u, &v);            add_Edge(u, v);            add_Edge(v, u);        }        solve();    }    return 0;}