• [POJ 3621] Sightseeing Cows


    [题目链接]

              http://poj.org/problem?id=3621

    [算法]

            01分数规划(最优比率环)

    [代码]

            

    #include <algorithm>  
    #include <bitset>  
    #include <cctype>  
    #include <cerrno>  
    #include <clocale>  
    #include <cmath>  
    #include <complex>  
    #include <cstdio>  
    #include <cstdlib>  
    #include <cstring>  
    #include <ctime>  
    #include <deque>  
    #include <exception>  
    #include <fstream>  
    #include <functional>  
    #include <limits>  
    #include <list>  
    #include <map>  
    #include <iomanip>  
    #include <ios>  
    #include <iosfwd>  
    #include <iostream>  
    #include <istream>  
    #include <ostream>  
    #include <queue>  
    #include <set>  
    #include <sstream>  
    #include <stdexcept>  
    #include <streambuf>  
    #include <string>  
    #include <utility>  
    #include <vector>  
    #include <cwchar>  
    #include <cwctype>  
    #include <stack>  
    #include <limits.h>
    using namespace std;
    #define MAXL 1010
    #define MAXP 5010 
    const double eps = 1e-4;
    const int T = 1e5;
    const int INF = 2e9;
    
    struct edge
    {
            int to,w,nxt;
    } e[MAXP];
    
    int i,L,P,a,b,w,tot;
    int f[MAXL],head[MAXL];
    double l,r,mid,ans;
    
    inline void addedge(int u,int v,int w)
    {
            tot++;
            e[tot] = (edge){v,w,head[u]};
            head[u] = tot;
    }
    inline bool spfa(double mid)
    {
            int i,cur,v;
            double w;
            static int cnt[MAXL];        
            static bool inq[MAXL];
            static double dist[MAXL];
            queue< int > q;
            memset(cnt,0,sizeof(cnt));
            memset(inq,false,sizeof(inq));
            while (!q.empty()) q.pop();
            for (i = 1; i <= L; i++)
            {
                    q.push(i);
                    cnt[i] = 1;
                    inq[i] = true;
                    dist[i] = -INF;
            }
            while (!q.empty())
            {
                    cur = q.front();
                    q.pop();
                    inq[cur] = false;
                    for (i = head[cur]; i; i = e[i].nxt)
                    {
                            v = e[i].to;
                            w = 1.0 * f[cur] - 1.0 * mid * e[i].w;
                            if (dist[cur] + w > dist[v])
                            {
                                    dist[v] = dist[cur] + w;
                                    if (!inq[v])
                                    {
                                            inq[v] = true;
                                            cnt[v]++;
                                            if (cnt[v] > L) return true;
                                            q.push(v);
                                    }
                            }
                    }
            }
            return false;
    }
    
    int main() 
    {
            
            scanf("%d%d",&L,&P);
            for (i = 1; i <= L; i++) scanf("%d",&f[i]);
            for (i = 1; i <= P; i++)
            {
                    scanf("%d%d%d",&a,&b,&w);
                    addedge(a,b,w);    
            }         
            l = 0; r = T;
            while (r - l > eps)
            {
                    mid = (l + r) / 2.0;
                    if (spfa(mid))
                    {
                            ans = mid;
                            l = mid;        
                    }    else r = mid;
            }    
            printf("%.2f
    ",ans);
            
            return 0;
        
    }

            

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  • 原文地址:https://www.cnblogs.com/evenbao/p/9391530.html
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