[NOI2012]迷失游乐园(期望,动态规划)
题目大意
给定一颗基环树或一颗树，从随机一个点出发，每次等概率走向之前没走过的点，知道不能走为止，求路径期望长度
数据范围
对于 (100\%) 的数据，(1leq W_ileq 100)。
| 测试点编号 | (n) | (m) | 备注 |
|---|---|---|---|
| (1) | (n=10) | (m = n-1) | 保证图是链状 |
| (2) | (n=100) | (m = n-1) | 只有节点 (1) 的度数大于 (2) |
| (3) | (n=1000) | (m = n-1) | / |
| (4) | (n=10^5) | (m = n-1) | / |
| (5) | (n=10^5) | (m = n-1) | / |
| (6) | (n=10) | (m = n) | / |
| (7) | (n=100) | (m = n) | 环中节点个数 (leq 5) |
| (8) | (n=1000) | (m = n) | 环中节点个数 (leq 10) |
| (9) | (n=10^5) | (m = n) | 环中节点个数 (leq 15) |
| (10) | (n=10^5) | (m = n) | 环中节点个数 (leq 20) |
解题思路
对于前 50% 部分--树
设 (e[x]) 为从 x 开始走的路径期望长度，(d[x]) 为从 x 开始一直向子树走的路径期望长度
(d[x]) 可以简单树形 dp 得到 (d[x] = sum_{y in subtreee(x)}frac {d[y]+w[x][y]}{son[x]})
e[x] 考虑换根 dp，维护从 x 向上走的期望长度 res，(res=frac{e[x]*(sons[x] + 1)-(d[y]+w[x][y])}{sons[x]} + w[x][y])，x 为根时要特殊讨论一下，(e[x] = frac{res + d[x]*sons[x]}{sons[x] + 1})，x 为根时 (e[x] = d[x])。
带码
void dfs1(int x, int fa) {
for (int i = h[x]; i; i = ne[i]) {
int y = to[i]; if (y == fa) continue;
dfs1(y, x); d[x] += d[y] + w[i], sons[x]++;
}
if (sons[x]) d[x] /= sons[x];
}
void dfs2(int x, int fa, double res) {
if (x != 1) e[x] = (d[x] * sons[x] + res) / (sons[x] + 1);
else e[x] = d[x];
for (int i = h[x]; i; i = ne[i]) {
int y = to[i]; if (y == fa) continue;
if (x != 1) dfs2(y, x, (e[x] * (sons[x] + 1) - (d[y] + w[i])) / sons[x] + w[i]);
else {
if (sons[x] > 1) dfs2(y, x, (e[x] * sons[x] - (d[y] + w[i])) / (sons[x] - 1) + w[i]);
else dfs2(y, x, w[i]);
}
}
}
对于剩下 50% 的数据--基环树
基环树首先应该做的就是把环单独考虑，我们可以想象成将一些有根树的根串起来成一个环，先对每棵树求一遍 d[x] 数组，再求出从每个根开始走环的期望路径长度，最后再换根 dp 从根转移到其他节点即可，细节：根的 sons要+=2
带码
const int N = 200500;
int h[N], ne[N<<1], to[N<<1], w[N<<1], tot = 1;
inline void add(int x, int y, int z) {
ne[++tot] = h[x], to[tot] = y, w[h[x] = tot] = z;
}
int vis[N], rt, Rt, sons[N], m, n;
double d[N], e[N];
void dfs1(int x, int fa) {
for (int i = h[x]; i; i = ne[i]) {
int y = to[i]; if (vis[y] || y == fa) continue;
dfs1(y, x); d[x] += d[y] + w[i], sons[x]++;
}
if (sons[x]) d[x] /= sons[x];
}
void dfs2(int x, int fa, double res) {
if (x != Rt) e[x] = (d[x] * sons[x] + res) / (sons[x] + 1);
for (int i = h[x]; i; i = ne[i]) {
int y = to[i]; if (y == fa || vis[y]) continue;
if (x != Rt) dfs2(y, x, (e[x] * (sons[x] + 1) - (d[y] + w[i])) / sons[x] + w[i]);
else {
if (sons[x] > 1) dfs2(y, x, (e[x] * sons[x] - (d[y] + w[i])) / (sons[x] - 1) + w[i]);
else dfs2(y, x, w[i]);
}
}
}
int To[N], Fr[N], wei[N];
void get_cir(int x, int fa) {
vis[x] = 1;
for (int i = h[x]; i; i = ne[i]) {
int y = to[i]; if (i == fa) continue;
if (rt) return; To[x] = y, Fr[y] = x, wei[y] = w[i];
if (vis[y]) rt = y; get_cir(y, i ^ 1);
}
}
void work(int x) {
double p = 1;
for (int nw = To[x]; nw != x; nw = To[nw]) {
e[x] += p * wei[nw], p /= sons[nw] + 1;
e[x] += p * sons[nw] * d[nw];
if (To[nw] == x) e[x] += p * d[nw];
}
p = 1;
for (int nw = Fr[x]; nw != x; nw = Fr[nw]) {
e[x] += p * wei[To[nw]], p /= sons[nw] + 1;
e[x] += p * sons[nw] * d[nw];
if (Fr[nw] == x) e[x] += p * d[nw];
}
e[x] = (e[x] + d[x] * sons[x]) / (sons[x] + 2);
sons[x] += 2, dfs2(Rt = x, 0, 0), sons[x] -= 2;
}
int main() {
read(n), read(m);
for (int i = 1, x, y, z; i <= m ; i++)
read(x), read(y), read(z), add(x, y, z), add(y, x, z);
if (m == n - 1) {
Rt = 1, dfs1(1, 0), e[1] = d[1], dfs2(1, 0, 0);
double ans = 0;
for (int i = 1;i <= n; i++) ans += e[i];
printf ("%.6lf
", ans / n);
return 0;
}
get_cir(1, 0); memset(vis, 0, sizeof(vis));
for (int x = rt; !vis[x]; x = To[x]) vis[x] = 1;
dfs1(rt, 0);
for (int x = To[rt]; x != rt; x = To[x]) dfs1(x, 0);
memset(vis, 0, sizeof(vis));
for (int x = rt; !vis[x]; x = To[x]) vis[x] = 1;
work(rt);
for (int x = To[rt]; x != rt; x = To[x]) work(x);
double ans = 0;
for (int i = 1;i <= n; i++) ans += e[i];
printf ("%.6lf
", ans / n);
return 0;
}