UVA

Problem  UVA - 11090 - Going in Cycle!!

Time Limit: 3000 mSec

Problem 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

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

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..
Constraints

• n ≤ 50

• a,b ≤ n

• c ≤ 10000000

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

题解：差分约束系统板子题，配上二分很容易解决，这里的spfa和一般的spfa稍有区别，原因我在UVA 11478的博客里解释过了，不再赘述，这种写法会比分别以每个点为源点跑高效不少，目前这份代码耗时50ms，但是分别以每个点为源点跑了200ms。

#include <bits/stdc++.h>

using namespace std;

#define REP(i, n) for (int i = 1; i <= (n); i++)
#define sqr(x) ((x) * (x))

const int maxn = 500 + 10;
const int maxm = 3000 + 10;
const int maxs = 10000 + 10;

typedef long long LL;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;

const LL unit = 1LL;
const int INF = 0x3f3f3f3f;
const LL mod = 1000000007;
const double eps = 1e-14;
const double inf = 1e15;
const double pi = acos(-1.0);

struct Edge
{
    int to, next;
    double w;
} edge[maxm << 1];

int n, m;
int tot, head[maxn];

void init()
{
    tot = 0;
    memset(head, -1, sizeof(head));
}

void AddEdge(int u, int v, double w)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    edge[tot].w = w;
    head[u] = tot++;
}

double dist[maxn];
bool vis[maxn];
int cnt[maxn];

bool spfa()
{
    queue<int> que;
    for (int i = 0; i < n; i++)
    {
        dist[i] = 0;
        cnt[i] = 0;
        vis[i] = true;
        que.push(i);
    }

    while (!que.empty())
    {
        int x = que.front();
        que.pop();
        vis[x] = false;
        for (int i = head[x]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if (dist[v] > dist[x] + edge[i].w)
            {
                dist[v] = dist[x] + edge[i].w;
                if (!vis[v])
                {
                    que.push(v);
                    vis[v] = true;
                    if (++cnt[v] > n)
                    {
                        return true;
                    }
                }
            }
        }
    }
    return false;
}

bool Judge(double x)
{
    for (int i = 0; i < tot; i++)
    {
        edge[i].w -= x;
    }
    bool ok = true;
    if(spfa())
        ok = false;
    for (int i = 0; i < tot; i++)
    {
        edge[i].w += x;
    }
    return ok;
}

int iCase;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int T;
    cin >> T;
    while (T--)
    {
        cin >> n >> m;
        init();
        int u, v;
        double w;
        double lim = -inf;
        for (int i = 0; i < m; i++)
        {
            cin >> u >> v >> w;
            u--, v--;
            AddEdge(u, v, w);
            lim = max(lim, w);
        }
        cout << "Case #" << ++iCase << ": ";
        if (Judge(lim + 1))
        {
            cout << "No cycle found." << endl;
        }
        else
        {
            double le = 0, ri = lim;
            while (ri - le > 1e-3)
            {
                double mid = (le + ri) / 2;
                if (Judge(mid))
                {
                    le = mid;
                }
                else
                {
                    ri = mid;
                }
            }
            cout << fixed << setprecision(2) << le << endl;
        }
    }
    return 0;
}