UVA11090 [Going in Cycle!!] 二分答案+SPFA判负圈

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题目链接

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题意：给你一个有向图，问你定义一个环的平均值为这个环上所有边的平均值，问你最小的环的平均值是多少。

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解题思路：我们可以[二分答案]，再用[SPFA]来check。

check方法：设最小环有k条边，Wi为第i条边的边权，猜测值为mid，那么有

(W1+W2+…+Wk) / k < mid

(W1+W2+…+Wk) < mid * k

(W1-mid)+(W2-mid)+…+(Wk-mid)<0

我们把每一条边权值都减去mid，再判断图中是否出现了负环。

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#include <queue> #include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#define Inf  100000000using namespace std;const int N = 55;struct Edge{    int u, v;    double dist;    Edge(int u, int v, double dist):u(u),v(v),dist(dist){}}; struct SPFA{    int n, m;    double d[N];    bool inq[N];    int cnt[N];    vector<Edge> edges;    vector<int> G[N];    void init(int n){        this->n=n;        for ( int i=1; i<=n; i++ ) G[i].clear();        edges.clear();    }     void addeage(int u, int v, double w){        edges.push_back(Edge(u,v,w));        m=edges.size();        G[u].push_back(m-1);    }    bool spfa(int s){        queue<int> Q;        for ( int i=1; i<=n; i++ ) d[i]=Inf, inq[i]=0, cnt[i]=0;        d[s]=0;        Q.push(s);        while( !Q.empty() ){            int u=Q.front(); Q.pop();            inq[u]=false;            for ( int i=0; i<G[u].size(); i++ ){                Edge& e=edges[G[u][i]];                if( d[e.v]>d[u]+e.dist ){                    d[e.v]=d[u]+e.dist;                    if( !inq[e.v] ){                        inq[e.v]=1;                        Q.push(e.v);                        if( ++cnt[e.v]==n ) return true;                    }                }            }        }        return false;       }}S;int n, m;bool binary(double x){    for ( int i=0; i<S.m; i++ ) S.edges[i].dist-=x;    bool cnt=false;    for ( int i=1; i<=n; i++ )         if( S.spfa(i) ){            cnt=true;            break;        }    for ( int i=0; i<S.m; i++ ) S.edges[i].dist+=x;    return cnt;}int main(){    int T;    scanf("%d", &T);    for ( int k=1; k<=T; k++ ){        scanf("%d%d", &n, &m );        S.init(n);        double ub;        for ( int i=1; i<=m; i++ ){            int u, v;            double w;            scanf("%d%d%lf", &u, &v, &w );            ub=max(ub,w);            S.addeage(u,v,w);        }        printf("Case #%d: ", k);        if( !binary(ub+1.0) ) printf("No cycle found.\n");        else {            double L=0, R=ub;            while( R-L>1e-3 ){                double M = L+(R-L)/2;                if( binary(M) ) R=M;                else L=M;            }            printf("%.2lf\n", L);        }     }}