uva11090 Going in Cycle!!

題意：找到圖中平均權值最小的迴路

思路：假設答案是val   那麼圖中的邊權值減去val，原來的迴路便會成爲負圈

val的值可以通過二分+spfa判負圈完成

//#pragma warning (disable: 4786)
//#pragma comment (linker, "/STACK:16777216")
//HEAD
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
//LOOP
#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
//STL
#define SZ(V) (int)V.size()
#define PB push_back
#define EQ(a, b) (fabs((a) - (b)) <= 1e-10)
#define ALL(c) (c).begin(), (c).end()
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RS(s) scanf("%s", s)
//OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(s) printf("%s\n", s)
typedef long long LL;
typedef unsigned long long ULL;
typedef vector <int> VI;
const int INF = 100000000;
const double eps = 1e-10;
const int MAXN = 505;


struct Edge{
    int from, to;
    double dist;
};

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});
        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] = 0;
            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] = G[u][i];
                    if (!inq[e.to])
                    {
                        Q.push(e.to);
                        inq[e.to] = 1;
                        if (++cnt[e.to] > n)
                            return true;
                    }
                }
            }
        }
        return false;
    }
}bell;

bool test(double mid)
{
    REP(i, bell.m)
        bell.edges[i].dist -= mid;
    bool ret = bell.negativeCycle();
    REP(i, bell.m)
        bell.edges[i].dist += mid;
    return ret;
}

int main()
{
    int n, m;
    int maxw;
    int T;
    RI(T);
    FE(kase, 1, T)
    {
        RII(n, m);
        bell.init(n);
        int a, b;
        int c;
        maxw = -INF;
        REP(i, m)
        {
            RIII(a, b, c);
            maxw = max(maxw, c);
            a--, b--;
            bell.addedge(a, b, c);
        }
        printf("Case #%d: ", kase);
        if (!test(maxw + 1))
            WS("No cycle found.");
        else
        {
            double L = 0.0, R = maxw;
            while (R - L > 1e-3)
            {
                double mid = (R + L) / 2;
                if (test(mid))
                    R = mid;
                else
                    L = mid;
            }
            printf("%.2lf\n", L);
        }
    }
}