• poj 1860 Currency Exchange (SPFA、正权回路 bellman-ford)


    链接:poj 1860

    题意:给定n中货币。以及它们之间的税率。A货币转化为B货币的公式为 B=(V-Cab)*Rab,当中V为A的货币量,

    货币S通过若干此转换,再转换为原本的货币时是否会添加

    分析:这个题就是推断是否存在正权回路。能够用bellman-ford算法,只是松弛条件相反

    也能够用SPFA算法,推断经过转换后,转换为原本货币的值是否比原值大、、、


    bellman-ford    0MS

    #include<stdio.h>
    #include<string.h>
    struct stu
    {
        int a,b;
        double r,c;
    }edge[205];
    double v,dis[105];
    int s;
    int bellmanford(int n,int m)
    {
        int i,j,flag=0;
        memset(dis,0,sizeof(dis));
        dis[s]=v;
        for(i=1;i<=n-1;i++)
            for(j=1;j<=m;j++)
                if(dis[edge[j].a]&&(dis[edge[j].a]-edge[j].c)*edge[j].r>dis[edge[j].b])
                    dis[edge[j].b]=(dis[edge[j].a]-edge[j].c)*edge[j].r;
        for(j=1;j<=m;j++)
            if(dis[edge[j].a]&&(dis[edge[j].a]-edge[j].c)*edge[j].r>dis[edge[j].b]){
                flag=1;
                break;
            }
        return flag;
    }
    int main()
    {
        int i,j,l,r,n,m,flag;
        while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF){
            j=1;
            for(i=1;i<=m;i++){
                scanf("%d%d",&l,&r);
                scanf("%lf%lf",&edge[j].r,&edge[j].c);
                edge[j].a=l;
                edge[j].b=r;
                j++;
                edge[j].a=r;
                edge[j].b=l;
                scanf("%lf%lf",&edge[j].r,&edge[j].c);
                j++;
            }
            flag=bellmanford(n,2*m);
            if(flag)
                printf("YES
    ");
            else
                printf("NO
    ");
        }
        return 0;
    }

    SPFA+邻接表  16MS

    #include<cstdio>
    #include<cstring>
    #include<queue>
    using namespace std;
    struct stu
    {
        int a,b;
        double r,c;
    }edge[205];
    double v,dis[105];
    int s,first[205],next[205],vis[105];
    int SPFA(int n,int m)
    {
        int i,pos;
        queue<int> q;
        memset(dis,0,sizeof(dis));
        memset(vis,0,sizeof(vis));
        dis[s]=v;
        q.push(s);
        vis[s]=1;
        while(!q.empty()){
            pos=q.front();
            q.pop();
            vis[pos]=0;
            i=first[pos];
            while(i!=-1){
                if((dis[pos]-edge[i].c)*edge[i].r>dis[edge[i].b]){
                    dis[edge[i].b]=(dis[pos]-edge[i].c)*edge[i].r;
                    if(!vis[edge[i].b]){
                        q.push(edge[i].b);
                        vis[edge[i].b]=1;
                    }
                }
                i=next[i];
            }
            if(dis[s]>v)
                return 1;
        }
        return 0;
    }
    int main()
    {
        int i,j,l,r,n,m,flag;
        while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF){
            j=1;
            for(i=1;i<=m;i++){
                scanf("%d%d",&l,&r);
                scanf("%lf%lf",&edge[j].r,&edge[j].c);
                edge[j].a=l;
                edge[j].b=r;
                j++;
                edge[j].a=r;
                edge[j].b=l;
                scanf("%lf%lf",&edge[j].r,&edge[j].c);
                j++;
            }
            memset(first,-1,sizeof(first));
            for(i=1;i<=2*m;i++){
                next[i]=first[edge[i].a];
                first[edge[i].a]=i;
            }
            flag=SPFA(n,2*m);
            if(flag)
                printf("YES
    ");
            else
                printf("NO
    ");
        }
        return 0;
    }




  • 相关阅读:
    sql server该账户当前被锁定,所以用户'sa'登录失败。系统管理员无法将该账户解锁。(Microsoft SQL Server,错误:18486)
    windows server常用操作
    sql server2005直接会话运行成功,但在作业执行报错
    sql server xp_readerrorlog引起的CPU爆满100%
    sql server错误日志
    (13)python网络编程,python Socket
    tcp/ip
    (12)python异常处理,python中的 try except
    典型分布式系统分析:Dynamo
    c++ set与unordered set的区别
  • 原文地址:https://www.cnblogs.com/zsychanpin/p/6971470.html
Copyright © 2020-2023  润新知