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. C# 客户端服务端的编写

    客户端的代码 class client { public void mehod() { TcpClient tcp = new TcpClient(); tcp.Connect(IPAddress.P ...

  2. 关于Dropdownlist使用的心得体会

    2013-07-23关于Dropdownlist使用的心得体会: Dropdownlist使用最多的几个属性: 一.Dropdownlist.Items,负责包含所有选项的容器 DropDownLis ...

  3. ultraedit比较两个文件差异经验

    链接地址:http://jingyan.baidu.com/article/fcb5aff7876551edab4a714b.html 程序开发人员经常要使用到两个文件的对比,有很多工具可以实现该功能 ...

  4. 怎样建立一个bower私库

    本教程适用于centos 安装之前 检查nodejs 假设没安装nodejs依照下面步骤安装 $ su - $ yum install openssl-devel $ cd /usr/local/sr ...

  5. 苹果2014WWDC亮点之个人浅见

    这届WWDC给人的整体感觉是融合.设备(手机IOS)和设备(电脑MAC OS X)的融合,人与信息的融合(SpotLight),人与代码的融合(Swift),人与人和设备的融合(HomeKit),接下 ...

  6. C#对数字添加逗号,千分位

    /// <summary> /// 对数字添加”,“号,可以处理负数以及带有小数的情况 /// </summary> /// <param name="vers ...

  7. QWidget类中默认是忽略inputMethodEvent事件(要获取输入的内容就必须使用这个事件)

    因为项目的需要以及主管的要求,准备将工程移植到Qt中,这样就可以比较容易的实现跨平台了.因为之前工程是在windows下开发的,第一个平台又是mobile所以除了底层框架之外其他的都是使用的windo ...

  8. CentOS 6.4 + 曙光DS200 IPSan组建FTP服务器

    CentOS 6.4 + 曙光DS200 IPSan组建FTP服务器 http://write.blog.csdn.net/postedit/10911105#本系列文章由ex_net(张建波)编写, ...

  9. on、where、having的区别(转载)

    on.where.having的区别 on.where.having这三个都可以加条件的子句中,on是最先执行,where次之,having最后.有时候如果这先后顺序不影响中间结果的话,那最终结果是相 ...

  10. Java Swing界面编程(22)---事件处理:动作事件及监听处理

    要想让一个button变得有意义,就必须使用事件处理.在swing的事件处理中.能够使用ActionListener接口处理button的动作事件. package com.beyole.util; ...