BZOJ 1778: [Usaco2010 Hol]Dotp 驱逐猪猡

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 const int mxn = 305;
 5 const int mxm = 50005;
 6 
 7 int n, m, p, q;
 8 
 9 double G[mxn][mxn];
10 
11 struct edge {
12     int x, y;
13 }E[mxm];
14 
15 int cnt[mxn];
16 
17 double K, F[mxn];
18 
19 signed main(void) {
20     scanf("%d%d%d%d", &n, &m, &p, &q);
21     
22     K = 1.0 * p / q;
23     
24     for (int i = 1; i <= m; ++i)
25         scanf("%d%d", &E[i].x, &E[i].y);
26     
27     for (int i = 1; i <= m; ++i)
28         ++cnt[E[i].x], ++cnt[E[i].y];
29     
30     for (int i = 1; i <= m; ++i)
31         G[E[i].x][E[i].y] -= (1.0 - K) / cnt[E[i].y],
32         G[E[i].y][E[i].x] -= (1.0 - K) / cnt[E[i].x];
33     
34     for (int i = 1; i <= n; ++i)
35         G[i][i] = 1.0;
36     
37     G[1][n + 1] = 1.0;
38     
39     for (int i = 1; i <= n; ++i)
40         for (int j = 1; j <= n; ++j)if (i != j){
41             double t = G[j][i] / G[i][i];
42             for (int k = 1; k <= n + 1; ++k)
43                 G[j][k] -= t * G[i][k];
44         }
45     
46     for (int i = 1; i <= n; ++i)
47         F[i] = G[i][n + 1] / G[i][i];
48     
49     for (int i = 1; i <= n; ++i)
50         printf("%.9lf\n", F[i] * K);
51 }