UVa 11090 在环中

https://vjudge.net/problem/UVA-11090

题意：

给定一个n个点m条边的加权有向图，求平均权值最小的回路。

思路：

二分枚举，只需要把每条边的权值变为w-mid，之后判断是否存在负圈即可。

#include <iostream>  
#include <cstring>  
#include <algorithm>   
#include <vector>
#include <queue>
using namespace std;

const int maxn = 50 + 5;

int n, m;

struct Edge
{
    int from, to;
    double dist;
    Edge(int u, int v, double d) :from(u), to(v), dist(d){}
};

struct BellmanFord
{
    int n, m;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool inq[maxn];
    double d[maxn];
    int p[maxn];
    int cnt[maxn];

    void init(int n)
    {
        this->n = n;
        for (int i = 0; i < n; i++)
            G[i].clear();
        edges.clear();
    }

    void AddEdge(int from, int to, double dist)
    {
        edges.push_back(Edge(from, to, dist));
        int m = edges.size();
        G[from].push_back(m - 1);
    }

    bool negativeCycle()
    {
        queue<int> Q;
        memset(inq, 0, sizeof(inq));
        memset(cnt, 0, sizeof(cnt));
        for (int i = 0; i < n; i++)
        {
            d[i] = 0; inq[0] = true; Q.push(i); 
        }

        while (!Q.empty())
        {
            int u = Q.front();
            Q.pop();
            inq[u] = false;
            for (int i = 0; i < G[u].size(); i++)
            {
                Edge& e = edges[G[u][i]];
                if (d[e.to]>d[u] + e.dist)
                {
                    d[e.to] = d[u] + e.dist;
                    p[e.to] = e.from;
                    if (!inq[e.to])
                    {
                        Q.push(e.to);
                        inq[e.to] = true;
                        if (++cnt[e.to] > n)   return true;
                    }
                }
            }
        }
        return false;
    }
}solver;

bool test(double x)
{
    for (int i = 0; i < m; i++)
        solver.edges[i].dist -= x;
    bool ret = solver.negativeCycle();
    for (int i = 0; i < m; i++)
        solver.edges[i].dist += x;
    return ret;
}

int main()
{
    //freopen("D:\input.txt", "r", stdin);
    int T;
    int u, v, d;
    scanf("%d", &T);
    for (int kase = 1; kase <= T; kase++)
    {
        scanf("%d%d", &n, &m);
        solver.init(n);
        int ub = 0;
        for (int i = 0; i < m; i++)    
        {
            scanf("%d%d%d", &u, &v, &d);
            ub = max(ub, d);
            solver.AddEdge(u - 1, v - 1, d);
        }
        printf("Case #%d: ", kase);
        if (!test(ub + 1))   printf("No cycle found.
");
        else
        {
            double L = 0, R = ub;
            while (R - L > 1e-3)
            {
                double M = L + (R - L) / 2;
                if (test(M))  R = M;
                else L = M;
            }
            printf("%.2lf
", L);
        }
    }
    return 0;
}