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\n", ans);
61     }
62     return 0;
63 }
View Code

 

posted on 2014-11-27 15:31  Xs酱~  阅读(206)  评论(0编辑  收藏  举报