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:
    APUE习题3.2用dup实现dup2以及shell中重定向符号的使用
    如何理解git checkout -- file和git reset HEAD -- file
    bash中通过设置PS1变量改变提示符颜色
    Ubuntu中root的默认密码
    Kali中装中文输入法小企鹅
    Find the Top 10 commands in your linux box!
    简明awk教程(Simple awk tutorial)
    PHP错误解决:Fatal error: Unknown: Failed opening required ...
    简单的端口扫描器(TCP connect)
    c# 爬虫(三) 文件上传
  • 原文地址:https://www.cnblogs.com/acjiumeng/p/6480467.html
Copyright © 2011-2022 走看看