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 =  + ;
 const int maxm = ;
 const int M =  ;

 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+) {
         que[i].clear();
         g[i].clear();
         vis[i] = false;
         REP(j, maxm) REP(k, maxm) {
             dp1[i][j][k] = , dp2[i][j][k] = ;
             if(j > k)
                 dp1[i][j][k] = ;
             if(j < k)
                 dp2[i][j][k] = ;
         }
     }
 }
 int n, m;
 short t1[maxm], t2[maxm];
 void tarjan(int u, int fa) {
     pa[u] = u;
     if(!fa) {
         Rep(i, , len+)
         if(s[i] == nd[u]) {
             dp1[u][i][i] = dp2[u][i][i] = ;
         }
     } else {
         int v = u;
         u = fa;
         Rep(i, , len+) Rep(j, , len+) {
             if(i >= j) {
                 dp2[v][i][j] = (dp2[v][i][j] + dp2[u][i][j] + (nd[v] == s[i])*(dp2[u][i-][j]))%M;
             }
             if(dp2[v][i][j] < )
                 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+][j]))%M;
             }
             if(dp1[v][i][j] < )
                 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\n", len ==  && s[] == nd[u]);
             continue;
         }
         SET(t1, );
         SET(t2, );
         t1[] = t2[len+] = ;
         if(w != u) {
             Rep(i, , len+) {
                 int tmp = ;
                 Rep(x, , i+) {
                     tmp = (tmp + ((LL)t1[x-]*dp1[w][x][i])%M)%M;
                 }
                 t1[i] = (dp1[u][][i]-tmp)%M;
                 if(t1[i] < )
                     t1[i] += M;
             }
         }
         if(w != v) {
             VREP(i, len, ) {
                 int tmp = ;
                 VREP(x, len, i) {
                     tmp = (tmp + ((LL)t2[x+]*dp2[w][x][i])%M)%M;
                 }
                 t2[i] = (dp2[v][len][i] - tmp)%M;
                 if(t2[i] < )
                     t2[i] += M;
             }
         }
         int ans = ;
         REP(i, len+) {
             if(s[i] == nd[w])
                 ans = (ans + ((LL)t1[i-]*t2[i+])%M)%M;
             ans = (ans + ((LL)t1[i]*t2[i+])%M)%M;
             if(ans < )
                 ans += M;
         }
         if(ans < )
             ans += M;
         printf("%d\n", ans);
     }
 }
 int main() {

     int T;
     for(int t = scanf("%d", &T); t <= T; t++) {
         scanf("%d%d", &n, &m);
         init(n);
         REP(i, n-) {
             int u, v;
             scanf("%d%d", &u, &v);
             g[u].pb(v);
             g[v].pb(u);
         }
         s[] = '\0';
         scanf("%s%s", nd+, s+);
         len = strlen(s+);
         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(, );
         solve();
     }
     return ;
 }