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 = ;
 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 = ; p <= n; ++p) {
         for (x = first[p]; x; x = e[x].next)
             a[p][y] = -(1.0 - P) / deg[y];
         a[p][p] = ;
     }
     a[][n + ] = P;
 }
 #undef y

 void gauss(int n) {
     int i, j, k;
     lf tmp;
     for (i = ; i <= n; ++i) {
         for (k = i, j = i + ; j <= n; ++j)
             if (fabs(a[j][i]) > fabs(a[k][i])) k = j;
         for (j = i; j <= n + ; ++j) swap(a[i][j], a[k][j]);
         for (k = i + ; k <= n; ++k)
             for (tmp = -a[k][i] / a[i][i], j = i; j <= n + ; ++j)
                 a[k][j] += a[i][j] * tmp;
     }
     for (i = n; i; --i) {
         for (j = i + ; j <= n; ++j)
             a[i][n + ] -= a[i][j] * ans[j];
         ans[i] = a[i][n + ] / a[i][i];
     }
 }

 int main() {
     int i, j;
     n = read(), m = read(), P = 1.0 * read() / read();
     for (i = ; i <= m; ++i)
         Add_Edges(read(), read());
     build_matrix();
     gauss(n);
     for (i = ; i <= n; ++i)
         printf("%.9lf\n", ans[i]);
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }