BZOJ1415（期望dp）

解法：

首先bfs预处理go数组：可可在j点时聪聪在点i是怎样贪心走的，这是为了之后O（1）获取转移线路。

然后dfs记忆化一下f[i][j]，代表从i到j的期望，对于每层：将所有情况的期望值相加。边界值是聪聪与可可在同一个点期望为0、聪聪一步或两步可到可可处期望为1。

 const int maxn = ;
 int n, m, st, ed;
 vector<int> dd[maxn];
 int go[maxn][maxn];
 db f[maxn][maxn];

 void bfs() {
     for (int i = ; i <= n; i++)
         sort(dd[i].begin(), dd[i].end());
     for (int i = ; i <= n; i++) {
         queue<P> Q;
         bool vis[n + ];
         memset(vis, false, sizeof vis);
         vis[i] = true;
         for (int t = ; t < dd[i].size(); t++) {
             int j = dd[i][t];
             Q.push(P(j, j));
             vis[j] = true;
             go[i][j] = j;
         }
         while (!Q.empty()) {
             P x = Q.front(); Q.pop();
             for (int t = ; t < dd[x.first].size(); t++) {
                 int j = dd[x.first][t];
                 if (!vis[j]) {
                     vis[j] = true;
                     go[i][j] = x.second;
                     Q.push(P(j, x.second));
                 }
             }
         }
     }
 }

 db dfs(int i, int j) {
     if (i == j)    return f[i][j] = ;
     if (go[i][j] == j || go[go[i][j]][j] == j)    return f[i][j] = ;
     if (f[i][j] == ) {
         db p = dd[j].size() + ;
         int nx = go[go[i][j]][j];
         for (int k = ; k < dd[j].size(); k++) {
             f[i][j] += dfs(nx, dd[j][k]);
         }
         f[i][j] += dfs(nx, j);
         f[i][j] = f[i][j] / p + ;
     }
     return f[i][j];
 }

 int main() {
     read(n), read(m), read(st), read(ed);
     for (int i = ; i <= m; i++) {
         int u, v;
         read(u), read(v);
         dd[u].push_back(v);
         dd[v].push_back(u);
     }
     bfs();
     printf("%.3lf\n", dfs(st, ed));
     return ;
 }