LOJ#2542 随机游走

解：首先minmax容斥变成经过集合t的第一个点就停止的期望步数。对于某个t，设从x开始的期望步数为f(x)

如果x∈t，f(x) = 0。否则f(x) = ∑f(y) / in[x] + 1

树上高斯消元。从叶子往上，可以发现每个点都可以表示为Af(fa) + B

于是我们推一波式子，参考，就可以对每个t，O(n)求出f(root)。

然后每个询问就枚举子集。

注意DFS的时候可以剪枝，遇到x∈t就返回，否则T飞.....

 #include <bits/stdc++.h>

 const int N = , MO = ;

 inline void read(int &x) {
     x = ;
     char c = getchar();
     while(c < '' || c > '') c = getchar();
     while(c >= '' && c <= '') {
         x = x *  + c - ;
         c = getchar();
     }
     return;
 }

 struct Edge {
     int nex, v;
 }edge[N << ]; int tp;

 int n, rt, e[N], A[N], B[N], now, in[N], ans[ << ], cnt[ << ], ans2[ << ];

 inline int qpow(int a, int b) {
     int ans = ;
     a = (a + MO) % MO;
     while(b) {
         if(b & ) ans = 1ll * ans * a % MO;
         a = 1ll * a * a % MO;
         b = b >> ;
     }
     return ans;
 }

 inline void add(int x, int y) {
     tp++;
     edge[tp].v = y;
     edge[tp].nex = e[x];
     e[x] = tp;
     return;
 }

 void DFS(int x, int f) {
     if((( << (x - )) | now) == now) {
         A[x] = B[x] = ;
         return;
     }
     int sa = , sb = ;
     for(int i = e[x]; i; i = edge[i].nex) {
         int y = edge[i].v;
         if(y == f) continue;
         DFS(y, x);
         sa = (sa + A[y]) % MO;
         sb = (sb + B[y]) % MO;
     }

     A[x] = qpow(in[x] - sa, MO - );
     B[x] = 1ll * A[x] * (in[x] + sb) % MO;

     return;
 }

 int main() {
     int q;
     read(n); read(q); read(rt);
     for(register int i = , x, y; i < n; i++) {
         read(x); read(y);
         add(x, y); add(y, x);
         in[x]++; in[y]++;
     }

     int lm =  << n;
     /*for(now = 1; now < lm; now++) {
         //memset(A, )
         DFS(rt, 0);
         ans[now] = B[rt];
     }*/
     memset(ans, -, sizeof(ans));
     memset(ans2, -, sizeof(ans2));

     for(register int i = ; i < lm; i++) {
         cnt[i] =  + cnt[i - (i & (-i))];
     }

     /// prework OVER
     for(register int i = , k; i <= q; i++) {
         read(k);
         int s = ;
         for(register int j = , x; j <= k; j++) {
             read(x);
             s |= ( << (x - ));
         }
         int Ans = ;
         if(ans2[s] != -) {
             Ans = ans2[s];
         }
         else {
             for(register int t = s; t; t = (t - ) & s) {
                 if(ans[t] == -) {
                     now = t;
                     DFS(rt, );
                     ans[t] = B[rt];
                 }
                 if(cnt[t] & ) Ans = (Ans + ans[t]) % MO;
                 else Ans = (Ans - ans[t] + MO) % MO;
             }
             ans2[s] = Ans;
         }
         printf("%d\n", (Ans + MO) % MO);
     }

     return ;
 }

AC代码