hdu 1245 Saving James Bond【floyd】

本文描述了一个基于电影情节的算法问题,即詹姆斯邦德如何在限定跳跃距离内逃离布满鳄鱼的湖泊。通过计算不同路径长度及所需跳跃次数,找到最短逃生路径。

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

Saving James Bond

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2954    Accepted Submission(s): 579

Problem Description

This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the center of a lake filled with crocodiles. There he performed the most daring action to escape -- he jumped onto the head of the nearest crocodile! Before the animal realized what was happening, James jumped again onto the next big head... Finally he reached the bank before the last crocodile could bite him (actually the stunt man was caught by the big mouth and barely escaped with his extra thick boot).
Assume that the lake is a 100×100 square one. Assume that the center of the lake is at (0,0) and the northeast corner at (50,50). The central island is a disk centered at (0,0) with the diameter of 15. A number of crocodiles are in the lake at various positions. Given the coordinates of each crocodile and the distance that James could jump, you must tell him whether he could escape.If he could,tell him the shortest length he has to jump and the min-steps he has to jump for shortest length.

Input

The input consists of several test cases. Each case starts with a line containing n <= 100, the number of crocodiles, and d > 0, the distance that James could jump. Then one line follows for each crocodile, containing the (x, y) location of the crocodile. Note that x and y are both integers, and no two crocodiles are staying at the same position. 

Output

For each test case, if James can escape, output in one line the shortest length he has to jump and the min-steps he has to jump for shortest length. If it is impossible for James to escape that way, simply ouput "can't be saved".

Sample Input

4 10

17 0

27 0

37 0

45 0

1 10

20 30

Sample Output 

42.50 5

can't be saved

Author

weigang Lee

 

题目大意:在直角坐标系中,以(-50,50)和(50,-50)为两个顶点组成的矩形表示这是要逃脱的区域,有一个以原点为中心,以15为直径的原型岛屿,主人公从岛屿出发,跳出矩形,问在每次跳跃不超过距离d条件下的:最小跳出的距离,和这种条件下的最小步数。


概括思路:

设定起点为(0,0),然后求出起点到其他各个点的最短距离,然后对于map【0】【i】,枚举其点i到终点的距离,然后加上map【0】【i】求最小的即可;


细节实现:

1、因为我们从起点出发其实并不是从(0,0)点出发跳跃,而是从以其围成的圆上的点出发开始跳跃的,所以:

map【0】【i】=dis(原点,i)-7.5;
map【i】【j】=dis(i,j);

2、对于求最短路部分:jumpp【i】【j】表示点i跳到点j需要跳跃几步。

因为点比较少,所以直接floyd即可,在floyd的过程的同时,其实就可以处理掉跳跃几步的问题:

if(map【j】【k】 >map【j】【i】+map【i】【k】)map【j】【k】 =map【j】【i】+map【i】【k】;jumpp【j】【k】=jumpp【j】【i】+jumpp【i】【k】;

这里注意一个点,如果map【j】【k】 ==map【j】【i】+map【i】【k】,我们要对jumpp【】【】进行单独更新,代码实现为:

        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                for(int k=0;k<=n;k++)
                {
                    if(map[j][k]>map[j][i]+map[i][k])
                    {
                        map[j][k]=map[j][i]+map[i][k];
                        jump[j][k]=jump[j][i]+jump[i][k];
                    }
                    if(fabs(map[j][k]-map[j][i]+map[i][k])<eps)//精度控制细节
                    {
                        jump[j][k]=min(jump[j][k],jump[j][i]+jump[i][k]);
                    }
                }
            }
        }
3、这样我们就能求出从起点到其他各点的距离,然后我们枚举所有map【0】【i】,并且判断从点i到终点的距离能否跳跃过去,如果能,相对比较大小即可,代码实现为:

        for(int i=0;i<=n;i++)
        {
            double dd=min(min(50-x[i],x[i]+50),min(50-y[i],y[i]+50));//点i到终点的距离。
            if(dd>d)continue;//如果跳不到终点继续判断下一个点。
            else
            {
                if(output>map[0][i]+dd)
                {
                    output=map[0][i]+dd;
                    jumpp=jump[0][i]+1;
                    ok=1;//标记上能够到达终点。
                }
                if(fabs(output-map[0][i]-dd)<eps)//这里不要忘记。
                {
                    jumpp=min(jumpp,jump[0][i]+1);
                }
            }
        }
4、注意一个点,如果d>42.5的话,就不需要这么些操作了,直接输出42.5 1即可。

AC代码:

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;
#define eps 1e-5
double map[105][105];
int jump[105][105];
int x[105];
int y[105];
double dis(int x1,int y1,int x2,int y2)
{
    double x=x1-x2;
    double y=y1-y2;
    return sqrt(x*x+y*y);
}
int main()
{
    int n;double d;
    while(~scanf("%d%lf",&n,&d))
    {
        x[0]=0;
        y[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&x[i],&y[i]);
        }
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                map[i][j]=0x3f3f3f3f;
                jump[i][j]=1;
                jump[i][i]=0;
            }
        }
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                if(i==j)
                {
                    map[i][j]=0;continue;
                }
                double dd=dis(x[i],y[i],x[j],y[j]);
                if(i==0)dd-=7.50;
                if(dd>d)continue;
                if(dd<0)continue;
                map[i][j]=min(dd,map[i][j]);
            }
        }
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                for(int k=0;k<=n;k++)
                {
                    if(map[j][k]>map[j][i]+map[i][k])
                    {
                        map[j][k]=map[j][i]+map[i][k];
                        jump[j][k]=jump[j][i]+jump[i][k];
                    }
                    if(fabs(map[j][k]-map[j][i]+map[i][k])<eps)
                    {
                        jump[j][k]=min(jump[j][k],jump[j][i]+jump[i][k]);
                    }
                }
            }
        }
        int ok=0;
        int jumpp=0;
        double output=0x3f3f3f3f;
        if(d>=42.50)
        {
            printf("42.50 1\n");
            continue;
        }
        for(int i=0;i<=n;i++)
        {
            double dd=min(min(50-x[i],x[i]+50),min(50-y[i],y[i]+50));
            if(dd>d)continue;
            else
            {
                if(output>map[0][i]+dd)
                {
                    output=map[0][i]+dd;
                    jumpp=jump[0][i]+1;
                    ok=1;
                }
                if(fabs(output-map[0][i]-dd)<eps)
                {
                    jumpp=min(jumpp,jump[0][i]+1);
                }
            }
        }
        if(ok==1)
        {
            printf("%.2lf %d\n",output,jumpp);
        }
        else printf("can't be saved\n");
    }
    return 0;
}
/*
12 5
8 0
9 0
10 0
13 0
18 0
23 0
28 0
33 0
38 0
39 0
43 0
48 0
16 5
8 0
9 0
12 0
17 0
22 0
27 0
32 0
31 0
3 0
45 0
13 0
19 0
25 0
37 0
42 0
47 0
*/




资源下载链接为: https://pan.quark.cn/s/f989b9092fc5 HttpServletRequestWrapper 是 Java Servlet API 中的一个工具类,位于 javax.servlet.http 包中,用于对 HttpServletRequest 对象进行封装,从而在 Web 应用中实现对 HTTP 请求的拦截、修改或增强等功能。通过继承该类并覆盖相关方法,开发者可以轻松地自定义请求处理逻辑,例如修改请求参数、添加请求头、记录日志等。 参数过滤:在请求到达处理器之前,可以对请求参数进行检查或修改,例如去除 URL 编码、过滤敏感信息或进行安全检查。 请求头操作:可以修改或添加请求头,比如设置自定义的 Content-Type 或添加认证信息。 请求属性扩展:在原始请求的基础上添加自定义属性,供后续处理使用。 日志记录:在处理请求前记录请求信息,如 URL、参数、请求头等,便于调试和监控。 跨域支持:通过添加 CORS 相关的响应头,允许来自不同源的请求。 HttpServletRequestWrapper 通过继承 HttpServletRequest 接口并重写其方法来实现功能。开发者可以在重写的方法中添加自定义逻辑,例如在获取参数时进行过滤,或在读取请求体时进行解密。当调用这些方法时,实际上是调用了包装器中的方法,从而实现了对原始请求的修改或增强。 以下是一个简单的示例,展示如何创建一个用于过滤请求参数的包装器: 在 doFilter 方法中,可以使用 CustomRequestWrapper 包装原始请求: 这样,每当调用 getParameterValues 方法时,都会先经过自定义的过滤逻辑。 HttpServletRequestWrapper 是 Java Web 开发中一个强大的工具,它提供了灵活的扩展性,允许开发者
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