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 = 1005;
int n, m, st, ed;
vector<int> dd[maxn];
int go[maxn][maxn];
db f[maxn][maxn];

void bfs() {
    for (int i = 1; i <= n; i++)
        sort(dd[i].begin(), dd[i].end());
    for (int i = 1; i <= n; i++) {
        queue<P> Q;
        bool vis[n + 1];
        memset(vis, false, sizeof vis);
        vis[i] = true;
        for (int t = 0; 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 = 0; 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] = 0;
    if (go[i][j] == j || go[go[i][j]][j] == j)    return f[i][j] = 1;
    if (f[i][j] == 0) {
        db p = dd[j].size() + 1;
        int nx = go[go[i][j]][j];
        for (int k = 0; 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 + 1;
    }
    return f[i][j];
}

int main() {
    read(n), read(m), read(st), read(ed);
    for (int i = 1; i <= m; i++) {
        int u, v;
        read(u), read(v);
        dd[u].push_back(v);
        dd[v].push_back(u);
    }
    bfs();
    printf("%.3lf
", dfs(st, ed));
    return 0;
}