bellman_ford寻找平均权值最小的回路

给定一个有向图，如果存在平均值最小的回路，输出平均值。

使用二分法求解，对于一个猜测值mid，判断是否存在平均值小于mid的回路

如果存在平均值小于mid的包含k条边的回路，那么有w1+w2+w3+...+wk < k * mid,即(w1-mid)+(w2-mid)+..(wk-mid)<0,

即判断是否存在负权回路即可。

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=&problem=2031&mosmsg=Submission+received+with+ID+14572787

#include <stdio.h>
#include <string.h>
const int N = 50+10;
const int INF = 1<<30;
struct Edge
{
    int u,v;
    double weight;
}g[4000000];
double dist[N];
void relax(int u, int v,double weight)
{
    if(dist[v] > dist[u] + weight)
        dist[v] = dist[u] + weight;
}
bool bellman_ford(int n, int m)
{
    int i,j;
    for(i=0; i<n-1; ++i)//n-1循环
        for(j=0; j<m; ++j)//枚举所有的边去松弛最短路径
        {
            relax(g[j].u,g[j].v,g[j].weight);
        }
    bool flag = false;
    for(i=0; i<m; ++i)
        if(dist[g[i].v] > dist[g[i].u] + g[i].weight)
        {
            flag = true;
            break;
        }
    return flag;
}
bool test(double x,int n, int m)
{
    int i;
    for(i=0; i<m; ++i)
        g[i].weight -= x;
    bool ret =  bellman_ford(n,m);
    for(i=0; i<m; ++i)
        g[i].weight += x;
    return ret;
}
int main()
{
    int n,m,i,t,tCase=1;
    double l,r;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=1; i<=n; ++i)
            dist[i] = INF;
        l = r = 0;
        for(i=0; i<m; ++i)
        {
            scanf("%d%d%lf",&g[i].u,&g[i].v,&g[i].weight);
            r = r > g[i].weight ? r : g[i].weight;
        }
        if(!test(r+1,n,m))printf("Case #%d: No cycle found.
",tCase++);
        else
        {
            double mid;
        
            while(r-l>0.001)//因为题目要求保留2位小数，所以当r-l>0.001时，l就是答案。
            {
                double mid = (r + l ) / 2;
                if(test(mid,n,m))
                    r = mid;
                else
                    l = mid;
            }
            printf("Case #%d: %.2lf
",tCase++,l);
        }
        
    }
    return 0;
}