Ultraviolet A 11090 Going in Cycle !! (Binary + spfa)

Source: Internet
Author: User

Returns a directed positive right edge graph with n vertices and m edges, and returns the loop with the smallest average value.

The binary mid value. Subtract the mid value from each edge to check whether a negative ring exists. But pay attention to this question... The source image is not necessarily strongly connected, so the general spfa cannot be used to determine the negative ring. To start the phase in spfa, each point must be added to the queue once. Make sure that each strongly connected component can be added.

 

#include<algorithm>#include<iostream>#include<cstring>#include<cstdlib>#include<fstream>#include<sstream>#include<bitset>#include<vector>#include<string>#include<cstdio>#include<cmath>#include<stack>#include<queue>#include<stack>#include<map>#include<set>#define FF(i, a, b) for(int i=a; i<b; i++)#define FD(i, a, b) for(int i=a; i>=b; i--)#define REP(i, n) for(int i=0; i<n; i++)#define CLR(a, b) memset(a, b, sizeof(a))#define debug puts("**debug**")#define LL long long#define PB push_backusing namespace std;const int maxn = 51;const double INF = 1000000000;struct Edge{    int from, to;    double dist;};int n, m;vector<Edge> edges;vector<int> G[maxn];bool inq[maxn];double d[maxn];int cnt[maxn];inline void init(){    REP(i, n) G[i].clear(); edges.clear();}void add(int a, int b, double c){    edges.PB((Edge){a, b, c});    int nc = edges.size();    G[a].PB(nc-1);}bool negacycle(){    queue<int> q;    CLR(inq, 0); CLR(cnt, 0);    REP(i, n)   d[i] = INF, q.push(i);    while(!q.empty())    {        int u = q.front(); q.pop();        inq[u] = false;        int nc = G[u].size();        REP(i, nc)        {            Edge e = edges[G[u][i]];            if(d[e.to] > d[u] + e.dist)            {                d[e.to] = d[u] + e.dist;                if(!inq[e.to])                {                    q.push(e.to);                    inq[e.to] = 1;                    if(++cnt[e.to] > n) return true;                }            }        }    }    return false;}bool ok(double m){    int nc = edges.size();    REP(i, nc) edges[i].dist -= m;    bool ret = negacycle();    REP(i, nc) edges[i].dist += m;    return ret;}int main(){    int T;    scanf("%d", &T);    FF(kase, 1, T+1)    {        scanf("%d%d", &n, &m);        init();        int a, b; double c;        double L = 0, R = 0, M;        while(m--)        {            scanf("%d%d%lf", &a, &b, &c); a--; b--;            add(a, b, c);            R = max(R, c);        }        printf("Case #%d: ", kase);        if(!ok(R + 1))        {            puts("No cycle found.");            continue;        }        while(R - L > 1e-3)        {            M = (L + R) / 2;            if(ok(M)) R = M;            else L = M;        }        printf("%.2lf\n", R);    }    return 0;}

 

Related Article

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.