Pet

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 472 Accepted Submission(s): 229

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
 
Recommend
liuyiding
应该是bfs dfs都可以过的吧,很水啦!
#include <iostream>
#include <stdio.h>
#include <vector>
#include <string.h>
using namespace std;
#define MAXN 100050
vector<int > vec[MAXN];
int visit[MAXN];
int dis,ans;
int dfs(int u,int d)
{
ans++;
visit[u]=1;
int i;
if(d>=dis)
return 1;
for(i=0;i<vec[u].size();i++)
{
if(!visit[vec[u][i]])
dfs(vec[u][i],d+1);
}
return 1;
}
int main()
{
int tcase,n,s,e,i;
scanf("%d",&tcase);
while(tcase--)
{
scanf("%d%d",&n,&dis);
for(i=0;i<n;i++)
vec[i].clear();
for(i=0;i<n-1;i++)
{
scanf("%d%d",&s,&e);
vec[s].push_back(e);
vec[e].push_back(s);
}
ans=0;
memset(visit,0,sizeof(visit));
dfs(0,0);
printf("%d\n",n-ans);
}
return 0;
}

hdu4707 Pet的更多相关文章

  1. HDU4707:Pet(DFS)

    Problem Description One day, Lin Ji wake up in the morning and found that his pethamster escaped. He ...

  2. UAT SIT QAS DEV PET

    UAT: User Acceptance Testing 用户验收测试SIT: System Integration Testing 系统集成测试PET: Performance Evaluation ...

  3. get a new level 25 battle pet in about an hour

    If you have 2 level 25 pets and any level 1 pet, obviously start with him in your lineup. Defeat all ...

  4. hduoj 4707 Pet 2013 ACM/ICPC Asia Regional Online —— Warmup

    http://acm.hdu.edu.cn/showproblem.php?pid=4707 Pet Time Limit: 4000/2000 MS (Java/Others)    Memory ...

  5. HDU 4707:Pet

    Pet Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submis ...

  6. Microsoft .NET Pet Shop 4

    Microsoft .NET Pet Shop 4:将 ASP.NET 1.1 应用程序迁移到 2.0 299(共 313)对本文的评价是有帮助 - 评价此主题 发布日期 : 2006-5-9 | 更 ...

  7. Pet

    Problem Description One day, Lin Ji wake up in the morning and found that his pethamster escaped. He ...

  8. asp.net的3个经典范例(ASP.NET Starter Kit ,Duwamish,NET Pet Shop)学习资料

    asp.net的3个经典范例(ASP.NET Starter Kit ,Duwamish,NET Pet Shop)学习资料 NET Pet Shop .NET Pet Shop是一个电子商务的实例, ...

  9. Pet(hdu 4707 BFS)

    Pet Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submiss ...

随机推荐

  1. python下的web服务器模块

    python下的web服务模块有三种: BaseHTTPServer: 提供基本的Web服务和处理器类,分别是HTTPServer和BaseHTTPRequestHandler SimpleHTTPS ...

  2. Currency System in Geraldion (Codeforces 560A)

    A  Currency System in Geraldion Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64 ...

  3. 浏览器与服务器间的交互(客服端 <---> 服务器)

    浏览器与服务器间的交互(客服端 <---> 服务器) 请求--->处理--->响应 对类HttpContext 内部成员的使用 例如 :Request .Response .  ...

  4. Codeforces 489A SwapSort

    这题第一次看的时候以为是区间替换,后来发现看错了,只是单纯的元素替换. 解题思路: 先对输入的序列加个数组排个序 遍历下来,如果和排序后的结果当前元素不同,设当前位置为 i, 则往下面找,设查找位置为 ...

  5. 开大Stack的一个小技巧

    在程序头部添加一行 #pragma comment(linker, "/STACK:16777216") 可有效开大堆栈 实验效果如下: 11330179 2014-08-05 1 ...

  6. ASP.NET MVC 5 学习教程:Details 和 Delete 方法详解

    原文 ASP.NET MVC 5 学习教程:Details 和 Delete 方法详解 在教程的这一部分,我们将研究一下自动生成的 Details 和Delete 方法. Details 方法 打开M ...

  7. springmvc中使用response的out.print问题

    public ModelAndView handleRequest(HttpServletRequest request, HttpServletResponse response) throws E ...

  8. 设计模式(八)装饰器模式Decorator(结构型)

    设计模式(八)装饰器模式Decorator(结构型) 1. 概述 若你从事过面向对象开发,实现给一个类或对象增加行为,使用继承机制,这是所有面向对象语言的一个基本特性.如果已经存在的一个类缺少某些方法 ...

  9. 【Unity 3D】学习笔记四十:射线

    射线 射线,类比的理解就是游戏中的子弹.是在3D世界里中一个点向一个方向发射的一条无终点的线.在发射的过程中,一旦与其它对象发生碰撞,就停止发射. 射线的原理 创建一个射线时,首先须要知道射线的起点和 ...

  10. 【转】增强 scite 编辑器的代码提示功能

    在 windows 下写 Lua, 我能找到的最好的编辑器就是 luaForWindows 项目里带的 scite. npp (即 notepad++ ) 也将就着能用, 不过只有代码高亮和简单的单词 ...