hdu 4601 Letter Tree

最蛋疼的是某节点相同字母的后续可能有多个,因此要将原树和字典树配合使用

rmq 算法

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <vector>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000") 
typedef long long ll;
const int MAXN = 200000;
const int MOD = 1000000007;
int n; 
ll f[25], mPow[MAXN];
struct node_edge
{
    int to, next, val;
}edge[MAXN<<1];
int head[MAXN], nmE;
void addEdge(int fr, int to, int ch)
{
    edge[nmE].to = to;
    edge[nmE].next = head[fr];
    edge[nmE].val = ch;
    head[fr] = nmE++;
    edge[nmE].to = fr;
    edge[nmE].next = head[to];
    edge[nmE].val = ch;
    head[to] = nmE++;
}
/// tire 
int c[MAXN][26]; ll val[MAXN];
int ntot;
/// end of tire
int newNode()
{
    const int sz = 26*4;
    memset(c[ntot], 0, sz);
    val[ntot] = 0;
    return ntot++;
}
struct Point
{
    int id, tim;
    bool operator < (const Point & a) const
    {
        return tim < a.tim;
    }
};
int step, dis[MAXN]/**记录每个节点想对于1节点的深度*/, tim[MAXN]/**对原树遍历的步骤,在区间上与tire遍历相同*/;
int pos[MAXN]/**原节点->tire中节点的映射关系*/; 
vector<Point> vec[MAXN]/**第i层的的节点*/;
int lon[MAXN]/**经过i节点所能构建的字符串最大长度*/, cnt[MAXN]/**从i节点往下的节点数量*/;
void dfs(int u, int fa, int cur)
{
    int i;
    Point tp;
    for (i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if (v == fa) continue;
        dis[v] = dis[u] + 1;
        tim[v] = ++step;
        tp.id = v, tp.tim = step;
        vec[dis[v]].push_back(tp);
        int k = edge[i].val;
        /**
            构建tire树
        */
        if (!c[cur][k])
        {
            c[cur][k] = newNode();
            val[c[cur][k]] = ((ll) val[cur]*26 % MOD + k)%MOD;
        }
        pos[v] = c[cur][k];
        dfs(v, u, c[cur][k]);
        lon[u] = max(lon[v], lon[u]);
        cnt[u] += cnt[v];
    }
    lon[u] = max(lon[u], dis[u]);
    ++cnt[u];
}
int rk[MAXN], g[MAXN];
void cal(int cur)
{
    rk[cur] = step++;
    g[rk[cur]] = cur;
    for (int i = 0; i< 26; ++i)
    {
        if (c[cur][i])
            cal(c[cur][i]);
    }
}
void init()
{
    for ( int i = 0; i<= n; ++i)
    {
        head[i] = -1; vec[i].clear();
        lon[i] = dis[i] = rk[i] = cnt[i] = 0;
    }
    ntot = nmE = 0;
    newNode();
}
int T, line[MAXN], dp[25][MAXN];
/// rmq算法,寻找区间的最大值
/**
    dp[i][j] = max(区间[j, j+2^i-1])
    dp[i][j] = max(dp[i-1][j], dp[i-1][j+2^(i-1)])
*/
void rmq()
{
    int i, j;
    T = 0;
    for ( i = 1; i<= n; ++i)
    {
        int l = vec[i].size();
        if (l == 0) break;
        line[i] = T + 1;
        for (j = 0; j< l; ++j)
            dp[0][++T] = rk[pos[vec[i][j].id]];
    }
    for (i = 1; f[i] <= T; ++i)
    {
        for (j = 1; j+f[i]-1 <= T; ++j)
        {
            dp[i][j] = max(dp[i-1][j], dp[i-1][j+f[i-1]]);
        } 
    }
}

int m_query(int l, int r)
{
    if (l== r) return dp[0][l];
    if (l > r) swap(l, r);
    int i, k;
    for (i = 0; i<= 22 ; ++i)
    {
        if (f[i] >= (r-l+1))
        {
            k = i-1;
            break;
        }
    }
    return max(dp[k][l], dp[k][r-f[k]+1]);
}

int main()   
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    int i, j, k, cs, q;
    f[0] = mPow[0] = 1;
    for (i = 1; i<= 22; ++i)
        f[i] = (ll) f[i-1]*2%MOD;
    for (i = 1; i< 100005; ++i)
        mPow[i] = (ll) mPow[i-1]*26%MOD;
    char str[5];
    scanf("%d", &cs);
    while (cs--)
    {
        scanf("%d", &n);
        init();
        for (i = 1; i< n; ++i)
        {
            scanf("%d%d%s", &j, &k, str);
            addEdge(j, k, str[0]-'a');
        }
        step = 0;
        dfs(1, -1, 0);
        step = 0;
        cal(0);
        rmq();
        pos[1] = 0;
        scanf("%d", &q);
        while (q--)
        {
            int u, m, dep;
            scanf("%d%d", &u, &m);
            if (m == 0)
            {
                puts("0");
                continue;
            }
            if (( dep = dis[u]+m) > lon[u])
            {
                puts("IMPOSSIBLE");
                continue;
            }
            Point tpm;
            tpm.tim = tim[u];
            int l = lower_bound(vec[dep].begin(), vec[dep].end(), tpm) - vec[dep].begin();
            if (l == vec[dep].size()) --l;
            l += line[dep];
            tpm.tim = tim[u] + cnt[u] - 1;
            int r = upper_bound(vec[dep].begin(), vec[dep].end(), tpm) - vec[dep].begin();
            --r;
            r += line[dep];
            int k = m_query(l, r);
            k = g[k];
            u = pos[u];
            printf("%I64d\n", (val[k]-val[u]*mPow[m]%MOD+MOD) % MOD);
        }
    }
    return 0;
}


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