洛谷 P2973 [USACO10HOL]Driving Out the Piggies G

洛谷传送门

思路

设 \(f_i\) 为经过第 \(i\) 个结点的期望次数，那么炸弹在第 \(i\) 个结点爆炸的概率即为 \(f_i \times \dfrac{P}{Q}\)。

有转移：

\[f_u = [u=1] + \sum\limits_{(u,v) \in E} \dfrac{1 - \frac{P}{Q}}{deg_v} \times f_v \]

其中 \(\dfrac{1 - \frac{P}{Q}}{deg_v}\) 表示从 \(v\) 的炸弹等概率选择一条出边运送且不引爆的概率。

高斯消元即可。时间复杂度 \(O(nm + n^3)\)。

代码

code

/*

p_b_p_b txdy
AThousandSuns txdy
Wu_Ren txdy
Appleblue17 txdy

*/

#include <bits/stdc++.h>
#define pb push_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 310;
const int maxm = 90000;

int n, m, p, q, head[maxn], len, deg[maxn];
ldb a[maxn][maxn];

struct edge {
	int to, next;
} edges[maxm];

void add_edge(int u, int v) {
	edges[++len].to = v;
	edges[len].next = head[u];
	head[u] = len;
}

void gauss() {
	for (int i = 1; i <= n; ++i) {
		int r = i;
		for (int j = i + 1; j <= n; ++j) {
			if (fabs(a[j][i]) > fabs(a[r][i])) {
				r = j;
			}
		}
		for (int j = i; j <= n + 1; ++j) {
			swap(a[i][j], a[r][j]);
		}
		for (int j = i + 1; j <= n + 1; ++j) {
			a[i][j] /= a[i][i];
		}
		a[i][i] = 1;
		for (int j = 1; j <= n; ++j) {
			if (i == j) {
				continue;
			}
			for (int k = i + 1; k <= n + 1; ++k) {
				a[j][k] -= a[j][i] * a[i][k];
			}
			a[j][i] = 0;
		}
	}
}

void solve() {
	scanf("%d%d%d%d", &n, &m, &p, &q);
	while (m--) {
		int u, v;
		scanf("%d%d", &u, &v);
		add_edge(u, v);
		add_edge(v, u);
		++deg[u];
		++deg[v];
	}
	ldb prob = (ldb)p / q;
	a[1][n + 1] = 1;
	for (int u = 1; u <= n; ++u) {
		a[u][u] = 1;
		for (int i = head[u]; i; i = edges[i].next) {
			int v = edges[i].to;
			a[u][v] -= (1 - prob) / deg[v];
		}
	}
	gauss();
	for (int i = 1; i <= n; ++i) {
		printf("%.10Lf\n", a[i][n + 1] * prob);
	}
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}