题目大意:

http://poj.org/problem?id=3621

题解:

首先我们容易发现最优的路线一定只经过一个环。
所以我们可以把点权合并到边权上.
然后就转化为了一个01分数规划问题
求解即可

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){x=0;char ch;bool flag = false;while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
const int maxn = 1024;
const int maxm = 5010;
const double eps = 1e-6;
inline int dcmp(const double &x){if(x < eps && x > -eps) return 0;return x > 0 ? 1 : -1;
}
struct Edge{int to,next,d;double dis;
}G[maxm<<1];
int head[maxn],cnt;
void add(int u,int v,int d){G[++cnt].to = v;G[cnt].next = head[u];head[u] = cnt;G[cnt].d = d;
}
double dis[maxn];
bool inq[maxn];
#define v G[i].to
bool dfs(int u){inq[u] = true;for(int i = head[u];i;i=G[i].next){if(dis[v] > dis[u] + G[i].dis){dis[v] = dis[u] + G[i].dis;if(inq[v]) return true;if(dfs(v)) return true;}}inq[u] = false;return false;
}
#undef v
int w[maxn],n,m;
inline bool check(double mid){memset(inq,0,sizeof inq);memset(dis,0,sizeof dis);for(int i=1;i<=cnt;++i)G[i].dis = mid*G[i].d - w[G[i].to];for(int i=1;i<=n;++i) if(dfs(i)) return true;return false;
}
int main(){read(n);read(m);for(int i=1;i<=n;++i) read(w[i]);for(int i=1,u,v,d;i<=m;++i){read(u);read(v);read(d);add(u,v,d);//add(v,u,d);}double l = .0,r = 1e9;while((r-l) > eps){double mid = (l+r)/2;if(check(mid)) l = mid;else r = mid;}printf("%.2lf\n",l);getchar();getchar();return 0;
}