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UVA11090 Going in Cycle!! （二分+SPFA判断有无负权）

2024-07-09 19:08:04   228人阅读

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

 


题意：

在一个有向图中找一个平均距离最小的环。


思路：

二分枚举平均最小距离，每次每条边减去这个距离，然后spfa（或者bellmanFord找负环）如果找到，说明平均最小距离比这个值要小，如果没找到，则说明平均最小距离比这个值大。

注意可能是不连通图~


代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define INF 0x3f3f3f3f
#define eps 1e-6
#define maxn 55
#define MAXN 10005
using namespace std;

int n,m,ans,cnt,sx;
double le,ri,mid,res;
bool vis[maxn];
double dist[maxn];
int head[maxn],num[maxn];
struct Node
{
    int v,next;
    double w;
}edge[MAXN];

void addedge(int u,int v,double w)
{
    cnt++;
    edge[cnt].v=v;
    edge[cnt].w=w;
    edge[cnt].next=head[u];
    head[u]=cnt;
}
bool SPFA(int k)
{
    int i,j,nx,v;
    memset(vis,0,sizeof(vis));
    memset(num,0,sizeof(num));
    for(i=1;i<=n;i++) dist[i]=INF;
    sx=k;
    queue<int>q;
    dist[sx]=0;
    vis[sx]=num[sx]=1;
    q.push(sx);
    while(!q.empty())
    {
        nx=q.front();
        vis[nx]=0;
        q.pop();
        for(i=head[nx];i;i=edge[i].next)
        {
            v=edge[i].v;
            if(dist[v]>dist[nx]+edge[i].w-mid)
            {
                dist[v]=dist[nx]+edge[i].w-mid;
                if(!vis[v])
                {
                    num[v]++;
                    if(num[v]>n) return true ;
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
    return false ;
}
int main()
{
    int i,j,u,v,t,test=0;
    double w;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        cnt=0;
        memset(head,0,sizeof(head));
        for(i=1;i<=m;i++)
        {
            scanf("%d%d%lf",&u,&v,&w);
            addedge(u,v,w);
        }
        le=0; ri=10000005;
        int flag;
        while(ri-le>eps)
        {
            mid=(le+ri)/2.0;
            flag=0;
            for(i=1;i<=n;i++)
            {
                if(SPFA(i))
                {
                    flag=1; break ;
                }
            }
            if(flag) ri=mid;
            else le=mid;
        }
        res=le;
        if(res<=10000001) printf("Case #%d: %.2f\n",++test,res);
        else printf("Case #%d: No cycle found.\n",++test);
    }
    return 0;
}





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