蒟蒻是此题第500个AC的233
毛线，一眼看上去很高端，还有个什么复杂的式子。。。
后来发现。。。是。。。Floyd水题
d[i][j]表示i、j之间的最短距离，cnt[i][j]表示i、j之间最短距离的条数
于是运用乘法原理更新cnt[i][j]即可
1 /************************************************************** 2 Problem: 1491 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:104 ms 7 Memory:936 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 13 using namespace std; 14 typedef long long ll; 15 typedef double lf; 16 const int N = 105; 17 18 int n, m; 19 int d[N][N]; 20 ll cnt[N][N]; 21 lf ans; 22 23 inline int read() { 24 int x = 0, sgn = 1; 25 char ch = getchar(); 26 while (ch < '0' || '9' < ch) { 27 if (ch == '-') sgn = -1; 28 ch = getchar(); 29 } 30 while ('0' <= ch && ch <= '9') { 31 x = x * 10 + ch - '0'; 32 ch = getchar(); 33 } 34 return sgn * x; 35 } 36 37 int main() { 38 int i, j, k, x, y, z; 39 n = read(), m = read(); 40 memset(d, 127/3, sizeof(d)); 41 for (i = 1; i <= m; ++i) { 42 x = read(), y = read(), z = read(); 43 d[x][y] = d[y][x] = z; 44 cnt[x][y] = cnt[y][x] = 1; 45 } 46 for (k = 1; k <= n; ++k) 47 for (i = 1; i <= n; ++i) if (i != k) 48 for (j = 1; j <= n; ++j) if (j != i && j != k) 49 if (d[i][j] > d[i][k] + d[k][j]) 50 d[i][j] = d[i][k] + d[k][j], 51 cnt[i][j] = cnt[i][k] * cnt[k][j]; 52 else if (d[i][j] == d[i][k] + d[k][j]) 53 cnt[i][j] += cnt[i][k] * cnt[k][j]; 54 for (k = 1; k <= n; ++k) { 55 ans = 0; 56 for (i = 1; i <= n; ++i) if (i != k) 57 for (j = 1; j <= n; ++j) if (j != i && j != k) 58 if (d[i][j] == d[i][k] + d[k][j]) 59 ans += (lf) (cnt[i][k] * cnt[k][j]) /cnt[i][j]; 60 printf("%.3lf ", ans); 61 } 62 return 0; 63 }