P2973 [USACO10HOL]Driving Out the Piggies G
简单的概率DP。$f _ i$ 为 $i$ 点爆炸的概率，按照题意列方程即可。
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| #include <cstdio>
#include <cmath>
#include <vector>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const int N = 310;
const int eps = 1e-12;
int n, m, p, q;
double A[N][N];
vector <int> g[N];
void Gauss()
{
for (int k = 0; k < n; k++)
{
int t = k;
for (int i = k + 1; i < n; i++)
fabs(A[i][k]) > fabs(A[t][k]) && (t = i);
if (fabs(A[t][k]) < eps)
continue;
for (int i = k; i <= n; i++)
swap(A[t][i], A[k][i]);
for (int i = n; i >= k; i--)
A[k][i] /= A[k][k];
for (int i = k + 1; i < n; i++)
for (int j = n; fabs(A[i][k]) > eps && j >= k; j--)
A[i][j] -= A[k][j] * A[i][k];
}
for (int i = n - 1; i >= 0; i--)
for (int j = i + 1; j < n; ++j)
A[i][n] -= A[i][j] * A[j][n];
}
int main ()
{
read(n), read(m), read(p), read(q);
double t = (double)p / q;
for (int a, b; m; m--)
{
read(a), read(b);
a--, b--;
g[a].push_back(b);
g[b].push_back(a);
}
for (int i = 0; i < n; i++)
{
A[i][i] = 1;
for (int j : g[i])
A[i][j] = -(1 - t) / g[j].size();
}
A[0][n] = t;
Gauss();
for (int i = 0; i < n; i++)
printf("%.9lf\n", A[i][n]);
return 0;
}
|