Pet

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 535 Accepted Submission(s): 258

Problem Description
One day, Lin Ji wake up in the morning and found that his pethamster escaped. He searched in the room but didn’t find the hamster. He tried to use some cheese to trap the hamster. He put the cheese trap in his room and waited for three days. Nothing but cockroaches was caught. He got the map of the school and foundthat there is no cyclic path and every location in the school can be reached from his room. The trap’s manual mention that the pet will always come back if it still in somewhere nearer than distance D. Your task is to help Lin Ji to find out how many possible locations the hamster may found given the map of the school. Assume that the hamster is still hiding in somewhere in the school and distance between each adjacent locations is always one distance unit.
 
Input
The input contains multiple test cases. Thefirst line is a positive integer T (0<T<=10), the number of test cases. For each test cases, the first line has two positive integer N (0<N<=100000) and D(0<D<N), separated by a single space. N is the number of locations in the school and D is the affective distance of the trap. The following N-1lines descripts the map, each has two integer x and y(0<=x,y<N), separated by a single space, meaning that x and y is adjacent in the map. Lin Ji’s room is always at location 0.

 
Output
For each test case, outputin a single line the number of possible locations in the school the hamster may be found.
 
Sample Input
1
10 2
0 1
0 2
0 3
1 4
1 5
2 6
3 7
4 8
6 9
 
Sample Output
2
 
Source
 

DFS(深度优先搜索)

import java.io.*;
import java.util.*;
public class Main {
BufferedReader bu;
PrintWriter pw;
int MAX=1000001;
int t,n,d,e,num; int head[]=new int[MAX];
Node node[]=new Node[MAX];
boolean boo[]=new boolean[MAX]; public static void main(String[] args) throws Exception{
new Main().work();
}
void work()throws Exception{
bu=new BufferedReader(new InputStreamReader(System.in));
pw=new PrintWriter(new OutputStreamWriter(System.out),true); t=Integer.parseInt(bu.readLine());
while(t--!=0){
String str[]; str=bu.readLine().split(" ");
n=Integer.parseInt(str[0]);
d=Integer.parseInt(str[1]);
e=0;
num=1;
Arrays.fill(head, -1);
Arrays.fill(boo,false); for(int i=1;i<n;i++){
str=bu.readLine().split(" ");
int a=Integer.parseInt(str[0]);
int b=Integer.parseInt(str[1]); add(a,b);
add(b,a);
}
boo[0]=true;
DFS(0,0);
pw.println(n-num);
}
} void DFS(int x,int index){
if(index>=d)
return;
for(int i=head[x];i!=-1;i=node[i].next){
int v=node[i].v;
if(!boo[v]){
boo[v]=true;
num++;
DFS(v,index+1);
}
}
} void add(int a,int b){
node[e]=new Node();
node[e].v=b;
node[e].next=head[a];
head[a]=e++;
} class Node{
int v;
int next;
}
}

BFS(广度优先搜索)

import java.io.*;
import java.util.*;
public class Main {
BufferedReader bu;
PrintWriter pw;
Queue<Integer> que;
int MAX=200010;
int t,n,d,e,num; int[] head=new int[MAX];
boolean[] boo=new boolean[MAX];
Node[] node=new Node[MAX];
int dis[]=new int[MAX];
public static void main(String[] args) throws Exception{
new Main().work();
}
void work()throws Exception{
bu=new BufferedReader(new InputStreamReader(System.in));
pw=new PrintWriter(new OutputStreamWriter(System.out),true);
que=new LinkedList<Integer>();
t=Integer.parseInt(bu.readLine());
String str[];
while(t--!=0){
str=bu.readLine().split(" ");
n=Integer.parseInt(str[0]);
d=Integer.parseInt(str[1]); Arrays.fill(head,-1);
Arrays.fill(boo, false);
Arrays.fill(dis,0);
que.clear(); e=0;
num=0; for(int i=1;i<n;i++){
str=bu.readLine().split(" ");
int a=Integer.parseInt(str[0]);
int b=Integer.parseInt(str[1]);
add(a,b);
add(b,a);
}
boo[0]=true;
que.add(0);
BFS();
pw.println(num);
}
}
void BFS(){
while(!que.isEmpty()){
int t=que.poll();
for(int i=head[t];i!=-1;i=node[i].next){
int u=node[i].v;
if(!boo[u]){
dis[u]=dis[t]+1;
boo[u]=true;
que.add(u);
}
}
}
for(int i=0;i<n;i++){
if(dis[i]>d&&boo[i])
num++;
}
}
void add(int a,int b){
node[e]=new Node();
node[e].v=b;
node[e].next=head[a];
head[a]=e++;
}
class Node{
int v;
int next;
}
}

HDU 4707 Pet(DFS(深度优先搜索)+BFS(广度优先搜索))的更多相关文章

  1. DFS_BFS(深度优先搜索 和 广度优先搜索)

    package com.rao.graph; import java.util.LinkedList; /** * @author Srao * @className BFS_DFS * @date ...

  2. 0算法基础学算法 搜索篇第二讲 BFS广度优先搜索的思想

    dfs前置知识: 递归链接:0基础算法基础学算法 第六弹 递归 - 球君 - 博客园 (cnblogs.com) dfs深度优先搜索:0基础学算法 搜索篇第一讲 深度优先搜索 - 球君 - 博客园 ( ...

  3. BFS广度优先搜索 poj1915

    Knight Moves Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 25909 Accepted: 12244 Descri ...

  4. 图的遍历BFS广度优先搜索

    图的遍历BFS广度优先搜索 1. 简介 BFS(Breadth First Search,广度优先搜索,又名宽度优先搜索),与深度优先算法在一个结点"死磕到底"的思维不同,广度优先 ...

  5. 算法竞赛——BFS广度优先搜索

    BFS 广度优先搜索:一层一层的搜索(类似于树的层次遍历) BFS基本框架 基本步骤: 初始状态(起点)加到队列里 while(队列不为空) 队头弹出 扩展队头元素(邻接节点入队) 最后队为空,结束 ...

  6. DFS+BFS(广度优先搜索弥补深度优先搜索遍历漏洞求合格条件总数)--09--DFS+BFS--蓝桥杯剪邮票

    题目描述 如下图, 有12张连在一起的12生肖的邮票.现在你要从中剪下5张来,要求必须是连着的.(仅仅连接一个角不算相连)  比如,下面两张图中,粉红色所示部分就是合格的剪取.  请你计算,一共有多少 ...

  7. 深度优先dfs与广度bfs优先搜索总结+例题

    DFS(Deep First Search)深度优先搜索 深度优先遍历(dfs)是对一个连通图进行遍历的算法.它的思想是从一个顶点开始,沿着一条路一直走到底,如果发现不能到达目标解,那就返回到上一个节 ...

  8. 步步为营(十六)搜索(二)BFS 广度优先搜索

    上一篇讲了DFS,那么与之相应的就是BFS.也就是 宽度优先遍历,又称广度优先搜索算法. 首先,让我们回顾一下什么是"深度": 更学术点的说法,能够看做"单位距离下,离起 ...

  9. 【js数据结构】图的深度优先搜索与广度优先搜索

    图类的构建 function Graph(v) {this.vertices = v;this.edges = 0;this.adj = []; for (var i = 0; i < this ...

随机推荐

  1. 你跟大牛之间仅仅差一个google

    google在中国被墙的厉害,http://209.116.186.231/ 这个地址能够訪问google.另外.有VPN或者某个奇妙的浏览器也能够. 非技术人员,还能够凑合着用百度,可是技术人员必须 ...

  2. (十六)JQuery Ready和angularJS controller的运行顺序问题

    项目中使用了JQuery和AngularJS框架,近期定位一个问题,原因就是JQuery Ready写在了angularJS controller之前,导致JQuery选择器无法选中须要的元素(由于a ...

  3. 关于php判断中文字符的问题

    在网上找了好多例子,还是这个靠谱点: UTF-8匹配: 在javascript中,要判断字符串是中文是很简单的.比如: var str = "php编程"; if (/^[\u4e ...

  4. U盘安装centos 7 提示 “Warning: /dev/root does not exist, could not boot” 解决办法

    1.查询磁盘 cd /dev ls 2.查询结果 sda 是我的硬盘对应的文件名(我机子只有一块硬盘),所以sda4就是U盘对应的文件名了,可以看到是sda4.至此我们重启一下,回到第一个图片所示的界 ...

  5. Python IDLE 运行错误:IDLE's subprocess didn't make connection. --已解决(原创)!

    Python IDLE 错误描述: Subprocess Startup ErrorIDLE's subprocess didn't make connection. Either IDLE can' ...

  6. rackup工具

    gem包rack提供了rackup工具来启动webapplication 下面是一个入门范例,使用 bundler 管理库的一个sinatra应用   在begin文件夹下有三个文件 begin.ru ...

  7. IE保护模式下ActiveX控件打不开共享内存的解决方案

    原文:http://www.cppblog.com/Streamlet/archive/2012/10/25/193831.html 感谢溪流漫话的投递 IE保护模式下,ActiveX控件会打不开别的 ...

  8. bottle-session 0.2 : Python Package Index

    bottle-session 0.2 : Python Package Index bottle-session 0.2 Download bottle-session-0.2.tar.gz Redi ...

  9. android 播放音乐-进度条

    今天学渣研究了一下使用MediaPlayer播放音乐时加入进度条,进度条如今用的是android自带的seekbar,后期会跟换UI的,在之前可以播放音乐的基础上,如今加入的主要功能有两个: 1实时显 ...

  10. CSDN-Code平台使用过程中的5点经验教训

    昨天又创建了一个项目,fucms,可是本地一直没有权限提交,搞了非常久,试了几十次,都不行,我是非常的灰心和郁闷.  刚刚,和CSDN-Code的官方客服咨询了非常久非常久,最终摸索出来了一些心得体会 ...