[NOI2012]迷失游乐园(期望,动态规划)

[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，\(1\leq W_i\leq 100\)。

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\n", 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\n", ans / n);
	return 0;
}