BZOJ 1491 社交网络 Floyd 最短路的数目

https://www.lydsy.com/JudgeOnline/problem.php?id=1491

 1 #include<bits/stdc++.h>
2 #define IOS ios::sync_with_stdio(false);//不可再使用scanf printf
3 #define Max(a, b) ((a) > (b) ? (a) : (b))//禁用于函数，会超时
4 #define Min(a, b) ((a) < (b) ? (a) : (b))
5 #define Mem(a) memset(a, 0, sizeof(a))
6 #define Dis(x, y, x1, y1) ((x - x1) * (x - x1) + (y - y1) * (y - y1))
7 #define MID(l, r) ((l) + ((r) - (l)) / 2)
8 #define lson ((o)<<1)
9 #define rson ((o)<<1|1)
10 #define Accepted 0
12 using namespace std;
14 {
15     int x=0,f=1;char ch=getchar();
16     while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
17     while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
18     return x*f;
19 }
20
21 typedef long long ll;
22 const int maxn = 100 + 10;
23 const int MOD = 1000000007;//const引用更快，宏定义也更快
24 const int INF = 1e9 + 7;
25 const double eps = 1e-6;
26
27 int Map[maxn][maxn];
28 ll p[maxn][maxn];
29 double ans[maxn];
30 int main()
31 {
32     int n, m;
33     scanf("%d%d", &n, &m);
34     for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)Map[i][j] = INF;
35     while(m--)
36     {
37         int u, v, w;
38         scanf("%d%d%d", &u, &v, &w);
39         Map[u][v] = Map[v][u] = w;
40         p[u][v] = p[v][u] = 1;//路径条数
41     }
42     for(int k = 1; k <= n; k++)
43     {
44         for(int i = 1; i <= n; i++)
45         {
46             for(int j = 1; j <= n; j++)
47             {
48                 if(Map[i][k] + Map[k][j] < Map[i][j])
49                 {
50                     Map[i][j] = Map[i][k] + Map[k][j];
51                     p[i][j] = p[i][k] * p[k][j];
52                 }
53                 else if(Map[i][k] + Map[k][j] == Map[i][j])
54                     p[i][j] += p[i][k] * p[k][j];
55             }
56         }
57     }
58     for(int k = 1; k <= n; k++)//k为s-t的必经点
59         for(int s = 1; s <= n; s++)
60             for(int t = 1; t <= n; t++)
61                 if(s != t && s != k && k != t)
62     {
63         if(Map[s][k] + Map[k][t] == Map[s][t])//最短路径可以经过k点
64         {
65             ans[k] += 1.0 * p[s][k] * p[k][t] / p[s][t];
66         }
67     }
68     for(int i = 1; i <= n; i++)printf("%.3f\n", ans[i]);
69     return Accepted;
70 }

posted @ 2018-09-20 20:59  _努力努力再努力x  阅读(218)  评论(0编辑  收藏  举报