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 = ;
 const double eps = 1e-;
 const int edge = ;
 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 = ;

 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,,sizeof(inq));
         memset(cnt,,sizeof(cnt));
         for(int i = ; i <= N; ++i) {
                 d[i] = ;
                 inq[i] = ;
                 q.push(i);
         }

         while(!q.empty()) {
                 int u = q.front(); q.pop();
                 inq[u] = ;
                 for(int e = first[u]; e != -; 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] ] = ;
                                         if(++cnt[ v[e] ] > N) return true;
                                 }
                         }
                 }
         }

         return false;

 }

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

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

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