【UVA11090】Going in Cycle!!

Going in Cycle!!

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• Description

You are given a weighted directed graph with n vertices and m edges. Each cycle in the graph has a weight, which equals to sum of its edges. There are so many cycles in the graph with different weights. In this problem we want to find a cycle with the minimum mean.

• Input Format

The first line of input gives the number of cases, N. N test cases follow. Each one starts with two numbers n and m. m lines follow, each has three positive number a, b, c which means there is an edge from vertex a to b with weight of c.

• Output Format

For each test case output one line containing Case #x: followed by a number that is the lowest mean cycle in graph with 2 digits after decimal place, if there is a cycle. Otherwise print No cycle found.

• Sample Input

2
2 1
1 2 1
2 2
1 2 2
2 1 3

• Sample Output

Case #1: No cycle found.
Case #2: 2.50

• Hint

n ≤ 50
a, b ≤ n
c ≤ 10000000

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• 题目大意

给定一个有n个顶点m条边的加权有向图，如果图中存在环（回路），环的平均值等于，环上边的权值之和除以构成环的边数，图中可能不止存在一个回路，计算平均权值最小的回路。

• 分析

这道题刚开始打的时候想法是从每个没被访问过点开始进行深搜，若搜到的点在本次搜索时已经被搜到过了，则认为发现一个环。每次搜索过的点都视为被访问过。但后来有神犇跟我说这样做不仅会超时还会少找到几个环（我也不知到为什么）。
正解：
这是一道最优比例环，属于01分数规划，即求一个环上某些信息的比值最优为多少。
显然答案满足单调性质，所以可以二分答案。对于一个二分出的Lim若有Σi∈Ew[i]size of E<Lim，则可转化为Σi∈Ew[i]−size of E∗Lim<0。
实际上就是把找出的环上的边权减去Lim，如果是个负环，则Lim合法。至于找负环用Spfa即可，如果一个点被入队超过n次，则这个点于一个负环上。

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#include <queue>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;queue <int> Q;const int N=55;const double Inf=1e15,Eps=1e-8;int last[N],Count[N],In[N],T,n,m,u,v,tot,Case;double Dist[N],L,R,Mid,w;struct Data{    int to,next;    double val;}E[N*N];void Addline(int u,int v,double w){    E[++tot].to=v; E[tot].next=last[u]; E[tot].val=w; last[u]=tot;}bool Check(double Lim){    for (int i=1;i<=n;i++) Count[i]=0,Dist[i]=Inf,In[i]=0;    for (;!Q.empty();) Q.pop();    for (int i=1;i<=n;i++)        if (!Count[i])            for (Q.push(i),In[i]=1,Count[i]=1,Dist[i]=0;!Q.empty();){                int u=Q.front();In[Q.front()]=0,Q.pop();                for (int j=last[u];j;j=E[j].next){                    int v=E[j].to;                    if (Dist[v]>Dist[u]+E[j].val-Lim){                        Dist[v]=Dist[u]+E[j].val-Lim;                        Count[v]=Count[u]+1;                        if (Count[v]>n) return true;                        if (In[v]) continue;                        In[v]=1; Q.push(v);                    }                }            }    return false;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    for (scanf("%d",&T);T;T--){        memset(last,0,sizeof(last));        tot=0;        Case++;        scanf("%d%d",&n,&m);        for (int i=1;i<=m;i++){            scanf("%d%d %lf",&u,&v,&w);            Addline(u,v,w);        }        for (L=0,R=Inf,Mid=(L+R)*0.5;R-L>=Eps;Mid=(L+R)*0.5){            if (Check(Mid)) R=Mid;            else L=Mid;        }        if (R==Inf) printf("Case #%d: No cycle found.\n",Case);        else printf("Case #%d: %.2f\n",Case,R);    }    //fclose(stdin); fclose(stdout);    return 0;}

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• 易错点

这题的数据可能存在自环，所以Spfa的时候取出队首后应马上把队首在队中的标记取消（据说模板就应该是这样的），否则会出现一些玄学的问题。今晚一整晚都在调这个问题，最后还是hjj大神把他的程序一点一点的改成我的样式，最后才找出来。我实在是太菜了！O_O