uva 11090

I I U P C 2 0 0 6
Problem G: Going in Cycle!!
Input: standard input Output: standard output
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	Output for Sample Input
2 2 1 1 2 1 2 2 1 2 2 2 1 3	Case #1: No cycle found. Case #2: 2.50
Problemsetter: Mohammad Tavakoli Ghinani Alternate Solution: Cho

 二分答案，判断是否有负权回路。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int MAX_N = 100;
const double eps = 1e-5;
const int edge = 3000;
int first[MAX_N],Next[edge],v[edge];
double w[edge];
bool inq[MAX_N];
int cnt[MAX_N];
double d[MAX_N];
int N,M;
double sum = 0;

void add_edge(int id,int u) {
        int e = first[u];
        Next[id] = e;
        first[u] = id;
}

bool bellman(double x) {
        queue<int> q;
        memset(inq,0,sizeof(inq));
        memset(cnt,0,sizeof(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 e = first[u]; e != -1; e = Next[e]) {
                        if(d[ v[e] ] > d[u] + w[e] - x) {
                                d[ v[e] ] = d[u] + w[e] - x;
                                if(!inq[ v[e] ]) {
                                        q.push( v[e] );
                                        inq[ v[e] ] = 1;
                                        if(++cnt[ v[e] ] > N) return true;
                                }
                        }
                }
        }

        return false;

}

void solve() {
        double l = 1,r = sum;
        while(r - l >= eps) {
                //printf("l = %f r = %f
",l,r);
                double mid = (l + r) / 2;
                if(bellman(mid)) r = mid;
                else l = mid;
        }
        if(bellman(sum + 1)) {
                printf("%.2f
",l);
        } else {
                printf("No cycle found.
");
        }
}

int main()
{
    //freopen("sw.in","r",stdin);
    int t;
    scanf("%d",&t);
    for(int ca = 1; ca <= t; ++ca) {
            scanf("%d%d",&N,&M);
            for(int i = 1; i <= N; ++i) first[i] = -1;
            sum = 0;
            for(int i = 0; i < M; ++i) {
                    int u;
                    scanf("%d%d%lf",&u,&v[i],&w[i]);
                    sum += w[i];
                    add_edge(i,u);
            }

            //printf("sum = %f
",sum);
            printf("Case #%d: ",ca);
            solve();
    }
    //cout << "Hello world!" << endl;
    return 0;
}