Solution

设现在走到了一个环，环内有 (n) 个点， (n) 条边，点权为 (f_{i})，边权为 (e_{i})
设 (k = sum_{i = 1}^{n}frac{f_{i}}{e_{i}})，显然是 0/1分数规划模型，变形可得： (sum_{i = 1}^{n}f_{i} - k * ei geq 0) 时 (k) 合法
此式不太好判断，我们在不等式两边乘上 (-1)，得 (sum_{i = 1}^{n}k * e_{i} - f_{i} leq 0) ，转换为图中负环的判定
若存在负环则此 (k) 合法
二分求解0/1分数规划即可

求负环的时候，初始距离全部设为 (0)
如果有负环的话此个距离 (0) 肯定能变得更小，赋0可以减少更新量，加快效率

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#define LL long long
#define REP(i, x, y) for(int i = (x);i <= (y);i++)
using namespace std;
int RD(){
    int out = 0,flag = 1;char c = getchar();
    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
    return flag * out;
    }
const int maxn = 2019,INF = 1e9, maxv = 20019;;
int head[maxn],nume = 1;
struct Node{
    int v,dis,nxt;
    }E[maxv << 3];
void add(int u,int v,int dis){
    E[++nume].nxt = head[u];
    E[nume].v = v;
    E[nume].dis = dis;
    head[u] = nume;
    }
int num, nr;
double f[maxn];//愉♂悦值
double d[maxn];
bool ins[maxn], vis[maxn], flag;
void SPFA_dfs(int u, double k){
	ins[u] = 1, vis[u] = 1;
	for(int i = head[u];i;i = E[i].nxt){
		int v = E[i].v;
		double dis = E[i].dis;
		if(d[u] + k * dis - f[v] <= d[v]){
			if(ins[v] || flag){flag = 1;return ;}
			d[v] = d[u] + k * dis - f[v];
			SPFA_dfs(v, k);
			}
		}
	ins[u] = 0;
	}
bool check(double k){
	REP(i, 1, num)d[i] = 0, vis[i] = 0, ins[i] = 0;
	flag = 0;
	REP(i, 1, num){
		if(!vis[i])d[i] = 0, SPFA_dfs(i, k);
		if(flag)return 1;
		}
	return flag;
	}
double search(double l, double r){
	double ans;
	while(r - l >= 1e-3){
		double mid = (l + r) / 2;
		if(check(mid))ans = mid, l = mid;
		else r = mid;
		}
	return ans;
	}
double maxx = 0;
int main(){
	num = RD(), nr = RD();
	REP(i, 1, num)f[i] = RD(), maxx += f[i];
	REP(i, 1, nr){
		int u = RD(), v = RD(), dis = RD();
		add(u, v, dis);
		}
	printf("%.2lf
", search(0, maxx));
	return 0;
	}