zoukankan      html  css  js  c++  java
  • poj2502最短路!

    have just moved from a quiet Waterloo neighbourhood to a big, noisy city. Instead of getting to ride your bike to school every day, you now get to walk and take the subway. Because you don't want to be late for class, you want to know how long it will take you to get to school.
    You walk at a speed of 10 km/h. The subway travels at 40 km/h. Assume that you are lucky, and whenever you arrive at a subway station, a train is there that you can board immediately. You may get on and off the subway any number of times, and you may switch between different subway lines if you wish. All subway lines go in both directions.Input

    Input consists of the x,y coordinates of your home and your school, followed by specifications of several subway lines. Each subway line consists of the non-negative integer x,y coordinates of each stop on the line, in order. You may assume the subway runs in a straight line between adjacent stops, and the coordinates represent an integral number of metres. Each line has at least two stops. The end of each subway line is followed by the dummy coordinate pair -1,-1. In total there are at most 200 subway stops in the city.

    Output

    Output is the number of minutes it will take you to get to school, rounded to the nearest minute, taking the fastest route.

    Sample Input

    0 0 10000 1000
    0 200 5000 200 7000 200 -1 -1 
    2000 600 5000 600 10000 600 -1 -1

    Sample Output

    21
    
    此题没输出,真是蛋疼,错了也不知道怎么改,最后还是参考得来,因为范围问题错了n次,最关键的地方就是储存图!!!
    处理时很要注重细节
    #include<iostream>
    #include<algorithm>
    #include<cstdio>
    #include<cmath>
    #include<queue>
    using namespace std;
    const double inf=1e30;
    double d[300],cost[300][300];
    int n=2;
    struct node{
       double x,y;
    }e[300];
    double dis(node a,node b)
    {
        return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
    }
    void dijkstra()
    {
        queue<int>Q;
        bool vis[300];
        for(int i=0;i<n;i++)
        {
            d[i]=inf;
            vis[i]=0;
        }
        d[0]=0;
        Q.push(0);
        vis[0]=1;
        while(!Q.empty()){
            int a=Q.front();
            Q.pop();
            vis[a]=0;
            for(int i=0;i<n;i++)
                if(d[i]>d[a]+cost[a][i])
                {
                    d[i]=d[a]+cost[a][i];
                    if(!vis[i])
                    {
                        vis[i]=1;
                        Q.push(i);
                    }
                }
        }
    }
    int main()
    {
        scanf("%lf%lf%lf%lf",&e[0].x,&e[0].y,&e[1].x,&e[1].y);
        for(int i=0;i<300;i++)
            for(int j=0;j<300;j++)
            {
                if(i==j)cost[i][j]=0;
                else cost[i][j]=inf;
            }
        double a,b;
        while(~scanf("%lf%lf",&a,&b)){
                int m=n;
                while(1){
                    if(a==-1&&b==-1)break;
                    e[n++]={a,b};
                    scanf("%lf%lf",&a,&b);
                }
            for(int i=m+1;i<n;i++)
            {
                cost[i][i-1]=cost[i-1][i]=dis(e[i],e[i-1])*3.0/2000;
            }
        }
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                cost[i][j]=min(cost[i][j],dis(e[i],e[j])*6.0/1000);
        dijkstra();
        printf("%.0f ",d[1]);
        return 0;
    }
  • 相关阅读:
    Asp.Net Mvc: 应用BindAttribute
    Mvc内建功能(DefaultModelBinder)自动绑定。
    生成随机字母字符串(数字字母混和)
    C#中实现输入汉字获取其拼音(汉字转拼音)的2种方法
    集合里查找数据
    C#自定义导出数据到Excel中的类封装
    MySQL性能优化的最佳20+条经验
    DevExpress.XtraGrid.view.gridview 属性说明
    c# 连接Mysql数据库
    ADO.NET 结构 集中数据库联接结构
  • 原文地址:https://www.cnblogs.com/acjiumeng/p/6480467.html
Copyright © 2011-2022 走看看