poj2253(dijistra变形+堆优化)

博客讲述青蛙Freddy想跳到Fiona所在石头,因距离超出跳跃范围需借助其他石头。定义两石头间青蛙距离为所有可能路径中所需最小跳跃范围。给出石头坐标,要求计算两石头间青蛙距离,此问题是找特殊定义下的最短路。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimming and instead reach her by jumping. 
Unfortunately Fiona's stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps. 
To execute a given sequence of jumps, a frog's jump range obviously must be at least as long as the longest jump occuring in the sequence. 
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones. 

You are given the coordinates of Freddy's stone, Fiona's stone and all other stones in the lake. Your job is to compute the frog distance between Freddy's and Fiona's stone. 

Input

The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy's stone, stone #2 is Fiona's stone, the other n-2 stones are unoccupied. There's a blank line following each test case. Input is terminated by a value of zero (0) for n. 

Output

For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

 

 

这个题是找最短路 但是最短路的定义是

对于一条从起点到终点的路径 该路径的长度为该路径中最长边的长度

我们要找到最短的路径

//
// Created by xingchaoyue on 2019/5/5.
//

#include<iostream>
#include<cstdio>
#include<queue>
#include<cmath>
using namespace std;
const int maxn = 2000;
const int inf = 0x3f3f3f3f;

double mp[maxn][maxn];
double dist[maxn];
int vis[maxn];
struct node{
    double x,y;
};
node p[maxn];
int n;
struct node1{
    int v;
    double d;
     node1(int _v,double _d):v(_v),d(_d){}
     bool operator <(const node1 b)const{
         return d>b.d;
     }
};

double getdis(node a,node b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void init(){
    for(int i =1;i<=n;++i){
        for(int j =1;j<=n;++j){
            mp[i][j] = inf;
        }
    }
}
void dijistra(int st,int n){
    for(int i =1;i<=n;++i){
        dist[i] = inf;
        vis[i] = 0;
    }
    dist[st] = 0;
    node1 temp(st,0);
    priority_queue<node1>q;
    q.push(temp);
    while(!q.empty()){
        node1 t = q.top();
        q.pop();
        int v = t.v;
        //cout<<v<<endl;
        double dis = t.d;
        if(vis[v])continue;
        vis[v] = 1;
        for(int i =1;i<=n;++i){
            //cout<<dist[i]<<" ";
            if(dist[i]>max(dist[v],mp[v][i]))
            {
                dist[i] = max(dist[v],mp[v][i]);
                node1 temp(i,dist[i]);
                q.push(temp);
            }
        }
        //cout<<endl;
    }

}
int main(){
    int t = 1;
    while(~scanf("%d",&n)&&n){
        init();
        for(int i =1;i<=n;++i){
            cin>>p[i].x>>p[i].y;
        }
        for(int i =1;i<=n;++i){
            for(int j =i+1;j<=n;++j){
                mp[i][j] = mp[j][i] = getdis(p[i],p[j]);
            }
        }
       /* for(int i =1;i<=n;++i){
            for(int j =1;j<=n;++j){
                cout<<mp[i][j]<<" ";
            }
            cout<<endl;
        }*/
        int st = 1;
        int ed = 2;
        dijistra(1,n);
        printf("Scenario #");
        cout<<t++<<endl;
        printf("Frog Distance = %.3f\n",dist[ed]);
        cout<<endl;

    }
}

 

内容概要:本文介绍了奕斯伟科技集团基于RISC-V架构开发的EAM2011芯片及其应用研究。EAM2011是一款高性能实时控制芯片,支持160MHz主频和AI算法,符合汽车电子AEC-Q100 Grade 2和ASIL-B安全标准。文章详细描述了芯片的关键特性、配套软件开发套件(SDK)和集成开发环境(IDE),以及基于该芯片的ESWINEBP3901开发板的硬件资源和接口配置。文中提供了详细的代码示例,涵盖时钟配置、GPIO控制、ADC采样、CAN通信、PWM输出及RTOS任务创建等功能实现。此外,还介绍了硬件申领流程、技术资料获取渠道及开发建议,帮助开发者高效启动基于EAM2011芯片的开发工作。 适合人群:具备嵌入式系统开发经验的研发人员,特别是对RISC-V架构感兴趣的工程师和技术爱好者。 使用场景及目标:①了解EAM2011芯片的特性和应用场景,如智能汽车、智能家居和工业控制;②掌握基于EAM2011芯片的开发板和芯片的硬件资源和接口配置;③学习如何实现基本的外设驱动,如GPIO、ADC、CAN、PWM等;④通过RTOS任务创建示例,理解多任务处理和实时系统的实现。 其他说明:开发者可以根据实际需求扩展这些基础功能。建议优先掌握《EAM2011参考手册》中的关键外设寄存器配置方法,这对底层驱动开发至关重要。同时,注意硬件申领的时效性和替代方案,确保开发工作的顺利进行。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值