HDU 4757 Tree (倍增算法求LCA + 可持久化Trie树)

来源：互联网发布：数码宝贝进化图鉴软件编辑：程序博客网时间：2024/06/05 05:19

题目大意：

就是现在给出一棵树, 结点个数不超过10W, 每个节点上有一个不超过2^16的非负整数, 然后10W次询问, 每次询问两个节点的路径上的所有数中异或上给出的数的最大值

大致思路：

刚开始做这个题想的是树链剖分之后用线段树套Trie树来做...结果悲剧地MLE了...

另外那样做得话复杂度其实是比较大的...每次询问都是16*(logn)*logn级别的..

后来发现是个可持久化Trie树...

表示第一次写可持久化Trie树, 感觉和主席树很像, 都是多个版本的Trie的集合体, 然后由要访问的版本的不同而从不同的根节点出发来访问

对于给出的树按照父亲和儿子的关系来建立可持久化Trie树

每次访问的时候访问到的Trie树版本是考虑了从当前结点到父亲结点的路径上的所有点的一颗Trie树, 那么对于每次的路径(u, v)的查询, 就是u开始的Trie与v开始的Trie的查询减去他们的LCA的查询, 然后利用贪心的思想求出可能的最大值即可

代码如下：

Result : Accepted Memory : 41812 KB Time : 1528 ms

/* * Author: Gatevin * Created Time:  2015/10/2 15:20:38 * 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(int _u, int _v, int _nex)    {        u = _u, v = _v, nex = _nex;    }    Edge(){}};Edge edge[maxn << 1];int head[maxn];int tot;void add_Edge(int u, int v){    edge[++tot] = Edge(u, v, head[u]);    head[u] = tot;}int w[maxn];int father[maxn][20];int dep[maxn];int n, m;/* * 可持久化Trie树 */struct Trie{#define maxnode 2800100    int next[maxnode][2];    int L;    int root[maxn];    int sz[maxnode];    void init()    {        L = 0;        memset(root, 0, sizeof(root));        memset(sz, 0, sizeof(sz));    }    int newnode()    {        next[L][0] = next[L][1] = -1;        L++;        return L - 1;    }        /*     * 一个以root[x]开始的Trie树是树上x向上到链的数组成的Trie树     * 然后树上每个节点都用sz[u][v]表示以u向上到根的值组成的Trie树中结点v代表的出现次数     */    void insert(int x, int fa, int value)//x的父亲节点是y    {        x = root[x];        int y = root[fa];        for(int i = 15; i >= 0; i--)        {            int nex = 0;            if(value & (1 << i)) nex = 1;            if(next[x][nex] == -1)            {                int id = newnode();                next[x][nex] = id;                next[x][nex^1] = next[y][nex^1];//合并                sz[next[x][nex]] = sz[next[y][nex]];            }            x = next[x][nex], y = next[y][nex];            ++sz[x];        }    }        int query(int x, int y, int z, int value)//z是x和y的lca    {        int ret = 0, ano = w[z];        x = root[x], y = root[y], z = root[z];        for(int i = 15; i >= 0; i--)        {            int nex = 0;            if(value & (1 << i)) nex = 1;            if(sz[next[x][nex ^ 1]] + sz[next[y][nex ^ 1]] - 2*sz[next[z][nex ^ 1]] > 0)//说明在[x, y]的链上有数提供这样的路走            {                x = next[x][nex ^ 1], y = next[y][nex ^ 1], z = next[z][nex ^ 1];                ret += (1 << i);            }            else            {                x = next[x][nex], y = next[y][nex], z = next[z][nex];            }           }        return max(ret, ano ^ value);    }};Trie trie;void dfs(int now){    trie.root[now] = trie.newnode();    trie.insert(now, father[now][0], w[now]);//按照树的列的顺序插入    for(int i = head[now]; i + 1; i = edge[i].nex)    {        if(edge[i].v == father[now][0]) continue;        father[edge[i].v][0] = now;        dep[edge[i].v] = dep[now] + 1;                dfs(edge[i].v);    }    return;}int swim(int x, int H){    for(int i = 0; H > 0; i++)    {        if(H & 1) x = father[x][i];        H >>= 1;    }    return x;}int lca(int x, int y){    int i;    if(dep[x] < dep[y]) swap(x, y);    int k = dep[x] - dep[y];    x = swim(x, k);    if(x == y) return x;    while(1)    {        for(i = 0; father[x][i] != father[y][i]; i++);        if(i == 0) return father[x][0];        x = father[x][i - 1];        y = father[y][i - 1];    }    return -1;}void answer(int x, int y, int z){    printf("%d\n", trie.query(x, y, lca(x, y), z));}bool get(int& x){    x = 0;    char c;    bool ined = 0;    while(!ined)    {        c = getchar();        if(c == EOF) break;        while(c >= '0' && c <= '9')        {            ined = 1;            x = (x << 3) + (x << 1) + c - '0';            c = getchar();        }    }    return ined;}int main(){    while(get(n))    {        get(m);        memset(head, -1, sizeof(head));        tot = 0;        for(int i = 1; i <= n; i++)            get(w[i]);        int u, v;        for(int i = 1; i < n; i++)        {            get(u); get(v);            add_Edge(u, v);            add_Edge(v, u);        }                trie.init();        memset(father, 0, sizeof(father));        dep[1] = 1;        father[1][0] = -1;        dfs(1);        for(int j = 1; (1 << j) <= n; j++)            for(int i = 1; i <= n; i++)                if(father[i][j - 1] != -1)                    father[i][j] = father[father[i][j - 1]][j - 1];//准备好求lca                while(m--)        {            int x, y, z;            scanf("%d %d %d", &x, &y, &z);            answer(x, y, z);        }    }    return 0;}

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