Just CS rookie practice on DFS\BFS. But details should be taken care of:

1. Ruby implementation got TLE so I switched to C++

2. There's one space after each output node, including the last one. But SPOJ doesn't clarify it clearly.

//

#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
using namespace std; typedef map<int, vector<int> > Graph; void bfs(Graph &g, int key)
{
vector<int> visited; visited.reserve(g.size());
for(unsigned ic = ; ic < g.size(); ic ++)
{
visited[ic] = ;
} queue<int> fifo;
fifo.push(key);
while(!fifo.empty())
{ int cKey = fifo.front();
fifo.pop(); if(visited[cKey - ] == )
{
cout << cKey <<" "; vector<int> &ajVec = g[cKey];
for(unsigned i = ; i < ajVec.size(); i ++)
{
fifo.push(ajVec[i]);
}
visited[cKey - ] = ;
}
} cout << endl;
} vector<int> dfs_rec;
void initDfsRec(Graph &g)
{
dfs_rec.clear();
unsigned gSize = g.size();
dfs_rec.reserve(gSize);
for(unsigned ic = ; ic < gSize; ic ++)
{
dfs_rec[ic] = ;
}
} void dfs(Graph &g, int key)
{
cout << key << " ";
dfs_rec[key - ] = ;
vector<int> &ajVec = g[key];
for(unsigned i = ; i < ajVec.size(); i ++)
{
if(dfs_rec[ajVec[i] - ] == )
{
dfs(g, ajVec[i]);
}
}
} int main()
{ int runcnt = ;
cin >> runcnt;
for(int i = ; i < runcnt; i ++)
{
Graph g; int nvert = ; cin >> nvert;
for(int n = ; n < nvert; n ++)
{
int vid = ; cin >> vid;
int cnt = ; cin >> cnt;
vector<int> ajVec;
for(int k = ; k < cnt; k ++)
{
int aj = ; cin >> aj;
ajVec.push_back(aj);
} g.insert(Graph::value_type(vid, ajVec));
} //
cout << "graph " << i + << endl; int ikey = , iMode = ;
cin >> ikey >> iMode;
while(!(ikey == && iMode == ))
{
if(iMode == )
{
initDfsRec(g);
dfs(g, ikey);
cout <<endl;
}
else if(iMode == )
{
bfs(g, ikey);
}
cin >> ikey >>iMode;
}
} return ;
}

SPOJ #442 Searching the Graph的更多相关文章

  1. Codeforces Round #236 (Div. 2) C. Searching for Graph(水构造)

    题目大意 我们说一个无向图是 p-interesting 当且仅当这个无向图满足如下条件: 1. 该图恰有 2 * n + p 条边 2. 该图没有自环和重边 3. 该图的任意一个包含 k 个节点的子 ...

  2. 构造图 Codeforces Round #236 (Div. 2) C. Searching for Graph

    题目地址 /* 题意:要你构造一个有2n+p条边的图,使得,每一个含k个结点子图中,最多有2*k+p条边 水得可以啊,每个点向另外的点连通,只要不和自己连,不重边就可以,正好2*n+p就结束:) */ ...

  3. C. Searching for Graph(cf)

    C. Searching for Graph time limit per test 1 second memory limit per test 256 megabytes input standa ...

  4. CF_402C Searching for Graph 乱搞题

    题目链接:http://codeforces.com/problemset/problem/402/C /**算法分析: 乱搞题,不明白题目想考什么 */ #include<bits/stdc+ ...

  5. Codeforces Round #236 (Div. 2)

    A. Nuts time limit per test:1 secondmemory limit per test:256 megabytesinput:standard inputoutput:st ...

  6. Reading task(Introduction to Algorithms. 2nd)

    Introduction to Algorithms 2nd ed. Cambridge, MA: MIT Press, 2001. ISBN: 9780262032933. Introduction ...

  7. apache atlas源码编译打包 centos

    参考:https://atlas.apache.org/InstallationSteps.html https://blog.csdn.net/lingbo229/article/details/8 ...

  8. Clone Graph leetcode java(DFS and BFS 基础)

    题目: Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors. ...

  9. SPOJ 375. Query on a tree (树链剖分)

    Query on a tree Time Limit: 5000ms Memory Limit: 262144KB   This problem will be judged on SPOJ. Ori ...

随机推荐

  1. PAT (Basic Level) Practise:1007. 素数对猜想

    [题目链接] 让我们定义 dn 为:dn = pn+1 - pn,其中 pi 是第i个素数.显然有 d1=1 且对于n>1有 dn 是偶数.“素数对猜想”认为“存在无穷多对相邻且差为2的素数”. ...

  2. PAT (Basic Level) Practise:1002. 写出这个数

    [题目链接] 读入一个自然数n,计算其各位数字之和,用汉语拼音写出和的每一位数字. 输入格式:每个测试输入包含1个测试用例,即给出自然数n的值.这里保证n小于10100. 输出格式:在一行内输出n的各 ...

  3. Java获得文件的创建时间(精确到秒)

    jni C/C++ 头文件:MyFileTime.h C/C++ code /* DO NOT EDIT THIS FILE - it is machine generated */#include ...

  4. Linux1.0源代码编译过程

    根据源代码包中的readme文件及http://chfj007.blog.163.com/blog/static/173145044201191195856806/?suggestedreading& ...

  5. 基于时间延迟的Python验证脚本

    自己写的一段Python脚本,经常拿来验证一些sqlmap等工具跑不出数据的网站. GET类型: import urllib import urllib2 import time payloads = ...

  6. urlscan使用详解

    0x01 简介与下载  URLScan是集成在IIS上的,可以制约的HTTP请求的安全工具.通过阻止特定的HTTP请求,URLScan安全工具有助于防止潜在的有害的请求到达服务器上的应用. 最新版UR ...

  7. Oracle语句

    分页查询: select rn,last_name,salary from( select rownum rn,last_name,salary from( select last_name,sala ...

  8. android中的DatePicker与TimePicker

    1.布局文件 <RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android" xmlns ...

  9. kuangbin_UnionFind D (HDU 3038)

    加权并查集 似乎就是在想这题的时候突然理解了之前看E题没看懂的标准加权解法 值得注意的技巧 为了让区间之前连成树 形式设定为为(l, r] 接受l的输入后先自减一下就可以了 #include < ...

  10. ExtJS4.2.1自定义主题(theme)样式详解

    (基于Ext JS 4.2.1版本) UI组件 学习ExtJS就是学习组件的使用.ExtJS4对框架进行了重构,其中最重要的就是形成了一个结构及层次分明的组件体系,由这些组件形成了Ext的控件. Ex ...