求平均值最小的环，如果平均值最小为x，则如果把每条边的权值都减(x+1)，那么新图将会有负环，用bellman ford判断。
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("d:\in1.txt","r",stdin); freopen("d:\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!=' ')return ch; } return EOF; } struct Edge { int from,to; double dist; }; const int maxn=55; vector<int> g[maxn]; vector<Edge> edge; double d[maxn]; int inq[maxn]; int inq_cnt[maxn]; int n,m; void init() { for(int i=1;i<=n;i++)g[i].clear(); edge.clear(); } void add(int u,int v,double w) { Edge e=(Edge){u,v,w}; edge.push_back(e); g[u].push_back(edge.size()-1); } bool negativeCycle(int s) { queue<int> q; memset(inq,0,sizeof(inq)); memset(inq_cnt,0,sizeof(inq_cnt)); for(int i=1;i<=n;i++) { d[i]=0; inq[i]=1; q.push(i); } while(!q.empty()) { int u=q.front();q.pop(); inq[u]=0; for(int i=0;i<g[u].size();i++) { Edge &e=edge[g[u][i]]; if(d[u]+e.dist<d[e.to]) { d[e.to]=d[u]+e.dist; if(!inq[e.to]) { inq[e.to]=1; q.push(e.to); if(++inq_cnt[e.to]>n)return true; } } } } return false; } bool check(double x) { for(int i=0;i<m;i++) edge[i].dist-=x; int flag=negativeCycle(1); for(int i=0;i<m;i++) edge[i].dist+=x; return flag; } int main() { int t; scanf("%d",&t); for(int ca=1;ca<=t;ca++) { scanf("%d%d",&n,&m); init(); for(int i=1;i<=m;i++) { int u,v; double w; scanf("%d%d%lf",&u,&v,&w); add(u,v,w); } double l=0,r=10000002; //printf("%d %d ",10000000,INF); while(r-l>10e-4) { double mid=l+(r-l)/2; if(!check(mid)) l=mid; else r=mid; } if(l>10000000) printf("Case #%d: No cycle found. ",ca); else printf("Case #%d: %.2lf ",ca,l); } return 0; }