BZOJ1778 [Usaco2010 Hol]Dotp 驱逐猪猡

首先我们列出转移矩阵$M$，$M_{i, j} = \frac {1 - \frac{p} {q}} {deg[i]}$(i,j之间有边)or $M_{i, j} = 0$(i,j之间没边)

则这个矩阵$M_{i, j}$表示的是站在某个点$i$，下一次走到$j$且没有爆炸的概率

我们再看$M^n_{i, j}$，表示的站在某个点$i$，走$n$步以后到达$j$且没有爆炸的概率

故$M^n$的第一列代表了$1$号点到其他所有点的概率，设为列向量$A_n$，则$A_n = M^n * B$，其中$B = (1, 0, 0, 0, ...)^T$

设第n步到各点且爆炸了的概率的列向量为$P_n$，则$P_n = \frac{p} {q} * A_n$

故答案列向量$Ans = \sum_{i = 0} ^ {+\infty} P_i$

把它展开：$Ans = \frac{p} {q} * (\sum_{i = 0} ^ {+\infty} M^i) * B$

由等比数列求和公式，$\sum_{i = 0} ^ {+\infty} M^i = \frac{I} {I - M} = (I - M)^{-1}$

故$Ans = \frac{p} {q} * (I - M)^{-1} * B$，即$(I- M) * Ans = \frac{p} {q} * B$

得到一个线性方程组，我们只要高斯消元即可

 

/**************************************************************
    Problem: 1778
    User: rausen
    Language: C++
    Result: Accepted
    Time:200 ms
    Memory:2264 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
#include <algorithm>
 
using namespace std;
typedef double lf;
const int N = 305;
const int M = N * N;
 
inline int read();
 
struct edge {
    int next, to;
    edge() {}
    edge(int _n, int _t) : next(_n), to(_t) {}
} e[M];
 
int n, m, deg[N];
int first[N], tot;
lf P, a[N][N], ans[N];
 
inline void Add_Edges(int x, int y) {
    e[++tot] = edge(first[x], y), first[x] = tot;
    e[++tot] = edge(first[y], x), first[y] = tot;
    ++deg[x], ++deg[y];
}
 
#define y e[x].to
inline void build_matrix() {
    int p, x;
    for (p = 1; p <= n; ++p) {
        for (x = first[p]; x; x = e[x].next)
            a[p][y] = -(1.0 - P) / deg[y];
        a[p][p] = 1;
    }
    a[1][n + 1] = P;
}
#undef y
 
void gauss(int n) {
    int i, j, k;
    lf tmp;
    for (i = 1; i <= n; ++i) {
        for (k = i, j = i + 1; j <= n; ++j)
            if (fabs(a[j][i]) > fabs(a[k][i])) k = j;
        for (j = i; j <= n + 1; ++j) swap(a[i][j], a[k][j]);
        for (k = i + 1; k <= n; ++k)
            for (tmp = -a[k][i] / a[i][i], j = i; j <= n + 1; ++j)
                a[k][j] += a[i][j] * tmp;
    }
    for (i = n; i; --i) {
        for (j = i + 1; j <= n; ++j)
            a[i][n + 1] -= a[i][j] * ans[j];
        ans[i] = a[i][n + 1] / a[i][i];
    }
}
 
int main() {
    int i, j;
    n = read(), m = read(), P = 1.0 * read() / read();
    for (i = 1; i <= m; ++i)
        Add_Edges(read(), read());
    build_matrix();
    gauss(n);
    for (i = 1; i <= n; ++i)
        printf("%.9lf\n", ans[i]);
    return 0;
}
 
inline int read() {
    static int x;
    static char ch;
    x = 0, ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}