HDOJ多校联合第六场

先一道一道题慢慢补上，

1009.题意，一棵N(N<=50000)个节点的树，每个节点上有一个字母值，给定一个串S0(|S0| <=30)，q个询问，(q<=50000)，每次询问经过两个点u，v之间的路径上的字母构成字符串S，问S0在S中作为序列出现了多少次。

分析：对于每次询问需要知道其LCA，设w = LCA(u, v),可以用tarjan处理出来，或者倍增法也行。然后w会将S0分成两部分，记做S1，S2，然后分别考虑S1在u->w的路径出现次数，以及S2在v->w出现的次数。

S1(x) = S0[1....x]，1<=x<=|S0|.

以S1为例，需要预处理出来u到根节点的路径上对应S0[i,j]序列出现的次数，设为dp1[u][i][j],然后S1(x)在u->w出现的次数记为t1[x],那么

t1[x] = dp1[u][1][x] - sum{t1[a-1]*dp1[fa[w]][a][x]} （1<=a<=x)

而dp1[u][1][x] 可以在tarjan求LCA时候预处理出来，转移dp1[u][i][j] = dp1[fa[u]][i][j] + (nd[u] == S0[i])*dp1[fa[u]][i+1][j];

对于S2也是可以同样考虑的。

注意：占用内存很大，需要用16位节省内存，dfs的话还需要扩栈，实现时发现不能随便用unsigned short,因为变成负数的话会相当于mod 2^16这个是会导致WA的。

代码：

#pragma comment(linker, "/STACK:16777216")
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <set>
#define in freopen("solve_in.txt", "r", stdin);
#define Rep(i, base, n) for(int i = (base); i < n; i++)
#define REP(i, n) for(int i = 0; i < (n); i++)
#define  VREP(i, n, base) for(int i = (n); i >= (base); i--)
#define SET(a, n) memset(a, (n), sizeof(a));
#define pb push_back
#define mp make_pair

using namespace std;
typedef vector<unsigned  short> VI;
typedef pair<unsigned  short, unsigned  short> PII;
typedef vector<PII> VII;
typedef long long LL;

const int maxn = 50000 + 10;
const int maxm = 33;
const int M = 10007 ;

short dp1[maxn][maxm][maxm], dp2[maxn][maxm][maxm];
bool vis[maxn];
int len;
char s[maxm], nd[maxn];
unsigned short pa[maxn];

VI g[maxn];
unsigned  short lca[maxn];
VII que[maxn];
VII qq;

unsigned  short findset(unsigned  short x) {
    return x == pa[x] ? x : pa[x] = findset(pa[x]);
}

void init(int n) {
    qq.clear();
    REP(i, n+1) {
        que[i].clear();
        g[i].clear();
        vis[i] = false;
        REP(j, maxm) REP(k, maxm) {
            dp1[i][j][k] = 0, dp2[i][j][k] = 0;
            if(j > k)
                dp1[i][j][k] = 1;
            if(j < k)
                dp2[i][j][k] = 1;
        }
    }
}
int n, m;
short t1[maxm], t2[maxm];
void tarjan(int u, int fa) {
    pa[u] = u;
    if(!fa) {
        Rep(i, 1, len+1)
        if(s[i] == nd[u]) {
            dp1[u][i][i] = dp2[u][i][i] = 1;
        }
    } else {
        int v = u;
        u = fa;
        Rep(i, 1, len+1) Rep(j, 1, len+1) {
            if(i >= j) {
                dp2[v][i][j] = (dp2[v][i][j] + dp2[u][i][j] + (nd[v] == s[i])*(dp2[u][i-1][j]))%M;
            }
            if(dp2[v][i][j] < 0)
                dp2[v][i][j] += M;
            if(i <= j) {
                dp1[v][i][j] = (dp1[v][i][j] + dp1[u][i][j] + (nd[v] == s[i])*(dp1[u][i+1][j]))%M;
            }
            if(dp1[v][i][j] < 0)
                dp1[v][i][j] += M;
        }
        u =v;
    }
    vis[u] = true;
    REP(i, g[u].size()) {
        int v = g[u][i];
        if(v == fa)
            continue;
        tarjan(v, u);
        pa[v] = u;
    }
    REP(i, que[u].size()) {
        int v = que[u][i].first;
        int id = que[u][i].second;
        if(!vis[v])
            continue;
        lca[id] = findset(v);
    }
}
void solve() {
    REP(ii, m) {
        int w = lca[ii];
        int u = qq[ii].first;
        int v = qq[ii].second;
        if(u == v) {
            printf("%d
", len == 1 && s[1] == nd[u]);
            continue;
        }
        SET(t1, 0);
        SET(t2, 0);
        t1[0] = t2[len+1] = 1;
        if(w != u) {
            Rep(i, 1, len+1) {
                int tmp = 0;
                Rep(x, 1, i+1) {
                    tmp = (tmp + ((LL)t1[x-1]*dp1[w][x][i])%M)%M;
                }
                t1[i] = (dp1[u][1][i]-tmp)%M;
                if(t1[i] < 0)
                    t1[i] += M;
            }
        }
        if(w != v) {
            VREP(i, len, 1) {
                int tmp = 0;
                VREP(x, len, i) {
                    tmp = (tmp + ((LL)t2[x+1]*dp2[w][x][i])%M)%M;
                }
                t2[i] = (dp2[v][len][i] - tmp)%M;
                if(t2[i] < 0)
                    t2[i] += M;
            }
        }
        int ans = 0;
        REP(i, len+1) {
            if(s[i] == nd[w])
                ans = (ans + ((LL)t1[i-1]*t2[i+1])%M)%M;
            ans = (ans + ((LL)t1[i]*t2[i+1])%M)%M;
            if(ans < 0)
                ans += M;
        }
        if(ans < 0)
            ans += M;
        printf("%d
", ans);
    }
}
int main() {
    
    int T;
    for(int t = scanf("%d", &T); t <= T; t++) {
        scanf("%d%d", &n, &m);
        init(n);
        REP(i, n-1) {
            int u, v;
            scanf("%d%d", &u, &v);
            g[u].pb(v);
            g[v].pb(u);
        }
        s[0] = '';
        scanf("%s%s", nd+1, s+1);
        len = strlen(s+1);
        REP(i, m) {
            int u, v;
            scanf("%d%d", &u, &v);
            qq.pb(mp(u, v));
            que[u].pb(mp(v, i));
            que[v].pb(mp(u, i));
        }
        tarjan(1, 0);
        solve();
    }
    return 0;
}