Uva 11090 最短路径

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2031

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

题目大意：给定n个点m条边的加权有向图，求平均权值最小的回路

大体思路：使用二分法求解，对于每一个猜测值mid，只需要判断是否存在平均值小于mid的回路。如何判断呢？假设存在一个包含K条边的回路，回路上个条边的权值为w1,w2

，，wk，那么平均值小于mid意味着w1+w2+w3+.....<K*mid即：(w1-mid)+(w2-mid)+(w3-mid)+....+(wk-mid)<0;换句话说，只要把每条边（a，b）的权值w(a,b)换成w(a,b)-mid,再判断是否有负权的回路即可。

#include<iostream>  #include<cstring>  #include<cstdio>  #include<string>  #include<algorithm>  #include<queue>    using namespace std;    #define eps 1e-8  #define INF 0x3f3f3f3f  #define MAXN 100    struct node  {      int to,next;      double dis;  }edge[MAXN*MAXN];    bool in[MAXN];  int head[MAXN],en;  int n,m,cnt[MAXN];  double dis[MAXN];    void add(int u,int v,double dis)  {      edge[en].to=v;      edge[en].dis=dis;      edge[en].next=head[u];      head[u]=en++;  }    bool spfa()  {      queue<int> q;      for(int i=1;i<=n;i++)      {          dis[i]=0;          cnt[i]=0;          in[i]=true;          q.push(i);      }      while(!q.empty())      {          int u=q.front();          in[u]=false;          q.pop();          for(int i=head[u];i!=-1;i=edge[i].next)          {              int v=edge[i].to;              if(dis[u]+edge[i].dis<dis[v])              {                  dis[v]=dis[u]+edge[i].dis;                  if(!in[v])                  {                      q.push(v);                      in[v]=true;                      if(++cnt[v]>=n)                          return false;                  }              }          }      }      return true;  }    bool jud(double x)  {      bool fg=0;      for(int i=1;i<=n;i++)          for(int j=head[i];j!=-1;j=edge[j].next)              edge[j].dis-=x;      if(!spfa()) fg=1;      for(int i=1;i<=n;i++)          for(int j=head[i];j!=-1;j=edge[j].next)              edge[j].dis+=x;      return fg;  }    int main()  {      int cs;      scanf("%d",&cs);      for(int t=1;t<=cs;t++)      {         int u,v;         double x;         double l=INF,r=0,mid;         scanf("%d%d",&n,&m);         memset(head,-1,sizeof(head));en=0;         for(int i=0;i<m;i++)         {             scanf("%d%d%lf",&u,&v,&x);             add(u,v,x);             l=min(l,x);             r=max(r,x);         }         printf("Case #%d: ",t);         if(!jud(r+1))         {              printf("No cycle found.\n");         }         else         {              while(r-l>eps)              {                  mid=l+(r-l)/2;                  if(jud(mid))                      r=mid;                  else                      l=mid;              }              printf("%.2lf\n",r);         }          }      return 0;  }