Saving James Bond - Hard Version

编程语言：Java
题目：

题解：路径还原罢了，每次跳到下一个点的时候记录一下就可以了，还以为能一次过，结果竟然可以一次跳到岸上（咱们强大的詹姆斯·邦德~~），也不想去优化程序了，直接一个外部条件判断解决；一个参数的distance是第一次跳的距离，后面的是判断两点之间的距离；此外，题目还有一个要求，在有很多最短路径的时候选择第一条最短的，解决方法就是按到岸的距离的远近来排序，这样还可以提前结束循环（到岸的距离太长了，一次跳不上去，那后面的也跳不上去了呀）；最后这道题用Scanner也不会超时的。
结果：AC

import java.io.*;
import java.util.*;
import java.util.concurrent.LinkedBlockingQueue;public class Main {static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));static Scanner sc = new Scanner(new BufferedInputStream(System.in));static crocodile[] c=new crocodile[102];static boolean[] f=new boolean[102];static int[] prev=new int[102];static double radius=7.5;static int n,d;public static void main(String[] args) throws IOException {in.nextToken();n= (int) in.nval;in.nextToken();d= (int) in.nval;for(int i=1;i<=n;i++){in.nextToken();int x= (int) in.nval;in.nextToken();int y= (int) in.nval;c[i]=new crocodile(x,y);}if(d>=50-radius)out.println(1);else{Arrays.sort(c,1,n+1, new Comparator<crocodile>() {@Overridepublic int compare(crocodile o1, crocodile o2) {return o1.x*o1.x+o1.y*o1.y-o2.x*o2.x-o2.y*o2.y;}});int ways=Integer.MAX_VALUE;Deque<Integer> stack=new LinkedList<>();for(int i=1;i<=n;i++){if(distance(i)){Arrays.fill(prev,-1);Arrays.fill(f,false);point p=bfs(i);if(p!=null&&p.step<ways){stack.clear();stack.push(p.i);while(prev[p.i]!=-1){stack.push(prev[p.i]);p.i=prev[p.i];}ways=p.step;}}else{break;}}if(ways==Integer.MAX_VALUE){out.println(0);}else{out.println(stack.size()+1);while(stack.size()>0){int i=stack.pop();out.println(c[i].x+" "+c[i].y);}}}out.flush();}private static point bfs(int i) {LinkedBlockingQueue<point> que=new LinkedBlockingQueue<>();que.add(new point(i,1));f[i]=true;while(!que.isEmpty()){point p=que.poll();if(Math.abs(c[p.i].x)>=50-d||Math.abs(c[p.i].y)>=50-d)return p;for(int j=1;j<=n;j++){if(!f[j]&&distance(p.i,j)){que.add(new point(j,p.step+1));prev[j]=p.i;f[j]=true;}}}return null;}private static boolean distance(int i) {if(Math.sqrt(c[i].x*c[i].x+c[i].y*c[i].y)<=d+radius)return true;elsereturn false;}private static boolean distance(int i, int j) {int x=c[i].x-c[j].x;int y=c[i].y-c[j].y;if(Math.sqrt(x*x+y*y)<=d)return true;elsereturn false;}}
class point{int i;int step;public point(int i, int step) {this.i = i;this.step = step;}
}
class crocodile{int x;int y;public crocodile(int x, int y) {this.x = x;this.y = y;}
}

Saving James Bond - Hard Version相关推荐

PTA 06-图2 Saving James Bond - Easy Version (25分)
题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/672 5-10 Saving James Bond - Easy Version ( ...
Saving James Bond - Easy Version 原创 2017年11月23日 13:07:33
06-图2 Saving James Bond - Easy Version(25 分) This time let us consider the situation in the movie &q ...
06-图2 Saving James Bond - Easy Version
题目来源:http://pta.patest.cn/pta/test/18/exam/4/question/625 This time let us consider the situation in ...
六、图(上):Saving James Bond - Easy Version
目录题目描述代码注意事项题目描述 This time let us consider the situation in the movie "Live and Let Die&quo ...
Saving James Bond - Easy Version
This time let us consider the situation in the movie "Live and Let Die" in which James Bon ...
中国大学MOOC-陈越、何钦铭-数据结构 Saving James Bond - Easy Version
题目描述: This time let us consider the situation in the movie "Live and Let Die" in which Jam ...
图5 Saving James Bond - Hard Version
全部每周作业和视频思考题答案和解析见浙江大学数据结构思考题+每周练习答案题目:This time let us consider the situation in the movie &qu ...
算法图2 Saving James Bond - Easy Version
全部每周作业和视频思考题答案和解析见浙江大学数据结构思考题+每周练习答案汇总题目:This time let us consider the situation in the movie & ...
(c语言)Saving James Bond - Hard Version (30分)
关于数据结构Mooc后的每一道答案基本我都已经给出了详解希望能对大家有所帮助收藏一下也是方便大家查找吧希望大家一起进步! (c语言)浙大数据结构Mooc作者答案集原题题目(谷歌翻译) 全检查 ...
07-图4. Saving James Bond - Hard Version (30)
本题测试点5是从小岛范围内可以直接跳到岸边-- 测试点4是验证步数第一跳最小的情况,刚开始没有考虑回溯,所以错了-- #include <stdio.h> #include <str ...

Saving James Bond - Hard Version

Saving James Bond - Hard Version相关推荐

最新文章

热门文章