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.....

 

  1 #include 
  2 
  3 const int N = 30, MO = 998244353;
  4 
  5 inline void read(int &x) {
  6     x = 0;
  7     char c = getchar();
  8     while(c < '0' || c > '9') c = getchar();
  9     while(c >= '0' && c <= '9') {
 10         x = x * 10 + c - 48;
 11         c = getchar();
 12     }
 13     return;
 14 }
 15 
 16 struct Edge {
 17     int nex, v;
 18 }edge[N << 1]; int tp;
 19 
 20 int n, rt, e[N], A[N], B[N], now, in[N], ans[1 << 19], cnt[1 << 19], ans2[1 << 19];
 21 
 22 inline int qpow(int a, int b) {
 23     int ans = 1;
 24     a = (a + MO) % MO;
 25     while(b) {
 26         if(b & 1) ans = 1ll * ans * a % MO;
 27         a = 1ll * a * a % MO;
 28         b = b >> 1;
 29     }
 30     return ans;
 31 }
 32 
 33 inline void add(int x, int y) {
 34     tp++;
 35     edge[tp].v = y;
 36     edge[tp].nex = e[x];
 37     e[x] = tp;
 38     return;
 39 }
 40 
 41 void DFS(int x, int f) {
 42     if(((1 << (x - 1)) | now) == now) {
 43         A[x] = B[x] = 0;
 44         return;
 45     }
 46     int sa = 0, sb = 0;
 47     for(int i = e[x]; i; i = edge[i].nex) {
 48         int y = edge[i].v;
 49         if(y == f) continue;
 50         DFS(y, x);
 51         sa = (sa + A[y]) % MO;
 52         sb = (sb + B[y]) % MO;
 53     }
 54 
 55     A[x] = qpow(in[x] - sa, MO - 2);
 56     B[x] = 1ll * A[x] * (in[x] + sb) % MO;
 57 
 58     return;
 59 }
 60 
 61 int main() {
 62     int q;
 63     read(n); read(q); read(rt);
 64     for(register int i = 1, x, y; i < n; i++) {
 65         read(x); read(y);
 66         add(x, y); add(y, x);
 67         in[x]++; in[y]++;
 68     }
 69 
 70     int lm = 1 << n;
 71     /*for(now = 1; now < lm; now++) {
 72         //memset(A, )
 73         DFS(rt, 0);
 74         ans[now] = B[rt];
 75     }*/
 76     memset(ans, -1, sizeof(ans));
 77     memset(ans2, -1, sizeof(ans2));
 78 
 79     for(register int i = 1; i < lm; i++) {
 80         cnt[i] = 1 + cnt[i - (i & (-i))];
 81     }
 82 
 83     /// prework OVER
 84     for(register int i = 1, k; i <= q; i++) {
 85         read(k);
 86         int s = 0;
 87         for(register int j = 1, x; j <= k; j++) {
 88             read(x);
 89             s |= (1 << (x - 1));
 90         }
 91         int Ans = 0;
 92         if(ans2[s] != -1) {
 93             Ans = ans2[s];
 94         }
 95         else {
 96             for(register int t = s; t; t = (t - 1) & s) {
 97                 if(ans[t] == -1) {
 98                     now = t;
 99                     DFS(rt, 0);
100                     ans[t] = B[rt];
101                 }
102                 if(cnt[t] & 1) Ans = (Ans + ans[t]) % MO;
103                 else Ans = (Ans - ans[t] + MO) % MO;
104             }
105             ans2[s] = Ans;
106         }
107         printf("%d\n", (Ans + MO) % MO);
108     }
109 
110     return 0;
111 }

AC代码

 

转载于:https://www.cnblogs.com/huyufeifei/p/10518598.html