Description

Input

第一行2個正整數，分別為n和m

以下m行，每行3個數，表示邊連接的信息，

Output

Sample Input

Sample Output

HINT

若設邊權為v，那麽n≤3000,m≤10000,v≤50000

題解

最小化平均值($01$分數規劃)。

使用二分求解。對於一個猜測的$mid$,只需判斷是否存在平均值小於$mid$的回路。

如何判斷?

假設存在一個包含$k$條邊的回路,回路上各邊權值為$w_1$ ,$w_2$ ,$...$,$w_k$ ,那麽平均值小於$midv意味著:

$$w_1 +w_2 +...+w_k <k×mid$$

即:

$$(w_1 -mid)+(w_2 -mid)+...+(w_k -mid)<0$$

換句話說,只要把邊$(a,b)$的權$w(a,b)$改成$w(a,b)-mid$,再判斷新圖中是否有負環即可。

存在負環,那麽之前的不等式滿足,即存在著更小的平均值,$r=mid$;不存在,$l=mid$。

 1 //It is made by Awson on 2017.10.9
 2 #include <set>
 3 #include <map>
 4 #include <cmath>
 5 #include <ctime>
 6 #include <cmath>
 7 #include <stack>
 8 #include <queue>
 9 #include <vector>
10 #include <string>
11 #include <cstdio>
12
 
 #include <cstdlib>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 #define LL long long
17 #define Min(a, b) ((a) < (b) ? (a) : (b))
18 #define Max(a, b) ((a) > (b) ? (a) : (b))
19 #define sqr(x) ((x)*(x))
20 using namespace std;
21 const double eps = 1e-9;
22 const int N = 3000;
23 const int M = 10000;
24 
25 int n, m, u, v, c;
26 bool vis[N+5];
27 double dist[N+5];
28 struct tt {
29   int to, next;
30   double cost;
31 }edge[M+5];
32 int path[N+5], top;
33 
34 void add(int u, int v, double c) {
35   edge[++top].to = v;
36   edge[top].next = path[u];
37   edge[top].cost = c;
38   path[u] = top;
39 }
40 bool dfs(int u, double dec) {
41   vis[u] = 1;
42   for (int i = path[u]; i; i = edge[i].next)
43     if (dist[edge[i].to] > dist[u]+edge[i].cost-dec) {
44       if (vis[edge[i].to]) return true;
45       dist[edge[i].to] = dist[u]+(double)edge[i].cost-dec;
46       if (dfs(edge[i].to, dec)) return true;
47     }
48   vis[u] = 0;
49   return false;
50 }
51 bool judge(double dec) {
52   for (int i = 1; i <= n; i++) {
53     memset(vis, 0, sizeof(vis));
54     memset(dist, 0, sizeof(dist));
55     if (dfs(i, dec)) return true;
56   }
57   return false;
58 }
59 void work() {
60   scanf("%d%d", &n, &m);
61   double L = 0, R = 0;
62   for (int i = 1; i <= m; i++) {
63     scanf("%d%d%d", &u, &v, &c);
64     add(u, v, c);
65     R = max(R, (double)c);
66   }
67   while (R-L >= eps) {
68     double mid = (L+R)/2.;
69     if (judge(mid)) R = mid;
70     else L =mid;
71   }
72   printf("%.8lf\n", (L+R)/2.);
73 }
74 int main() {
75   work();
76   return 0;
77 }

[HNOI 2009]最小圈