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 tireint 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;}