一开始不理解为什么不能直接用(Tarjan)跑出换直接求出最小值，然后想到了“简单环”，恍然大悟。
二分答案，把所有边都减去(mid)，判是否存在负环，存在就(r=mid)。
别的题都卡dfs，这题卡bfs
#include <cstdio>
#include <queue>
#define INF 2147483647
using namespace std;
const int MAXN = 30010;
const int MAXM = 100010;
const double eps = 1e-10;
struct Edge{
int next, to;
double dis;
}e[MAXM];
int num, head[MAXN];
inline void Add(int from, int to, double dis){
e[++num] = (Edge){ head[from], to, dis }; head[from] = num;
}
int a[MAXM], b[MAXN], T, n, m, v[MAXN], now;
double dis[MAXN], c[MAXN];
queue <int> q;
/*int bfs(int x){
while(q.size()) q.pop();
q.push(x); cnt[x] = 1; v[x] = 1; dis[x] = 0;
while(q.size()){
now = q.front(); q.pop(); v[now] = 0;
for(int i = head[now]; i; i = e[i].next)
if(dis[e[i].to] > dis[now] + e[i].dis){
dis[e[i].to] = dis[now] + e[i].dis;
if(!v[e[i].to]){
if(++cnt[e[i].to] >= n) return 1;
v[e[i].to] = 1;
q.push(e[i].to);
}
}
}
return 0;
}*/
int dfs(int u){
v[u] = 1;
for(int i = head[u]; i; i = e[i].next)
if(dis[e[i].to] > dis[u] + e[i].dis){
dis[e[i].to] = dis[u] + e[i].dis;
if(v[e[i].to]) return 1;
if(dfs(e[i].to)) return 1;
}
v[u] = 0;
return 0;
}
int check(){
for(int i = 1; i <= n; ++i)
if(dfs(i))
return 1;
return 0;
}
double l = -1e7, r = 1e7, mid;
int main(){
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; ++i)
scanf("%d%d%lf", &a[i], &b[i], &c[i]);
while(r - l > eps){
mid = (l + r) / 2;
for(int i = 1; i <= m; ++i)
Add(a[i], b[i], c[i] - mid);
if(check()) r = mid;
else l = mid;
for(int i = 1; i <= n; ++i)
head[i] = v[i] = dis[i] = 0;
num = 0;
}
printf("%.8lf
", l);
return 0;
}