UVA 11090 Going in Cycle!! 二分答案 + bellman-ford

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
#elsefreopen("d:\\in1.txt","r",stdin);freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch() {int ch;while((ch=getchar())!=EOF) {if(ch!=' '&&ch!='\n')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\n%d\n",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.\n",ca);elseprintf("Case #%d: %.2lf\n",ca,l);}return 0;
}