讓我看到你們的雙手
Going in Cycle!!
Description

You are given a weighted directed graph with n vertices and m edges. Each cycle in the graph has aweight, which equals to sum of its edges. There are so many cycles in the graph with different weights.
In this problem we want to nd a cycle with the minimum mean.

Input

The rst line of input gives the number of cases, N. N test cases follow. Each one starts with twonumbers n and m. m lines follow, each has three positive number a; b; c which means there is an edgefrom vertex a to b with weight of c.

Output

For each test case output one line containing Case #x: followed by a number that is the lowest meancycle in graph with 2 digits after decimal place, if there is a cycle. Otherwise print No cycle found..
Constraints
• n <= 50
• a; b <= n
• c <= 10000000

Sample Input

2
2 1
1 2 1
2 2
1 2 2
2 1 3

Sample Output

Case #1: No cycle found.
Case #2: 2.50

【解題報告】
題目大意是給定一個有向圖，求平均值最小的迴路。
二分答案w，然後每條邊都減去w判斷是否有負環。
（有誰能告訴我為什麼我的程式碼那麼慢。。。）

程式碼如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<deque>
using namespace
 
 std;
#define N 60
#define inf 0x3f3f3f3f
#define eps 1e-3

int head[N],vis[N],cnt_e[N];
double dis[N];
int t,n,m,cnt;
struct Edge
{
    int to,nxt;
    double w;
}e[N*N];

void adde(int u,int v,double w)
{
    e[++cnt].w=w;
    e[cnt].to=v;
    e[cnt].nxt=head[u];
    head[u]=cnt;
}
bool SPFA(int s)
{
    for(int i=0;i<=n;++i) dis[i]=inf*1.0; 
    memset(vis,0,sizeof(vis));
    memset(cnt_e,0,sizeof(cnt_e));
    deque<int> q;
    dis[s]=0,vis[s]=1;
    q.push_back(s);
    while(!q.empty())
    {
        int now=q.front();q.pop_front();
        vis[now]=0;
        for(int i=head[now];~i;i=e[i].nxt)
        {
            int v=e[i].to;
            if(dis[v]>dis[now]+e[i].w)
            {
                dis[v]=dis[now]+e[i].w;
                if(!vis[v])
                {
                    vis[v]=1;
                    if(!q.empty())
                    {
                        if(dis[v]>dis[q.front()]) q.push_back(v);
                        else q.push_front(v);
                    }
                    else q.push_back(v);
                    if(++cnt_e[v]>=n) return 0;
                }
            }
        }       
    }
    return 1;
}
bool check(double x)
{
    bool flag=0;
    for(int i=1;i<=n;++i)
    for(int j=head[i];~j;j=e[j].nxt) e[j].w-=x;
    for(int i=1;i<=n;++i) if(!SPFA(i)) flag=1;
    for(int i=1;i<=n;++i)
    for(int j=head[i];~j;j=e[j].nxt) e[j].w+=x;
    return flag;
}

int main()
{
    int cas=0;
    for(scanf("%d",&t);t;--t)
    {
        printf("Case #%d: ",++cas);
        double l=inf*1.0,r=0.0,mid=0.00;
        memset(head,-1,sizeof(head));
        cnt=0;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;++i)
        {
            int u,v;double w;
            scanf("%d%d%lf",&u,&v,&w);
            adde(u,v,w);
            l=min(l,w);
            r=max(r,w);
        }
        if(!check(r+1)) {printf("No cycle found.\n");continue;}
        while(r-l>eps)
        {
            mid=(r+l)/2.0;
            check(mid)?r=mid:l=mid;
        }
        printf("%.2lf\n",r);
    }
    return 0;
}