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

最新推荐文章于 2020-02-12 20:07:57 发布

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最新推荐文章于 2020-02-12 20:07:57 发布

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本文介绍了一种通过二分搜索和SPFA算法寻找有向图中平均边权重最小的环的方法。通过不断调整候选平均距离并检测是否存在负权环来确定最小平均距离。

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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）假设找到，说明平均最小距离比这个值要小。假设没找到。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;
}