<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[仰望圣光的plusplus7]]></title><description><![CDATA[愿圣光永远照耀着你]]></description><link>https://blog.youkuaiyun.com/sssogs</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; sssogs]]></copyright><item><title><![CDATA[POJ 2849 brainf*ck]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/17379913</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/17379913</guid><author>sssogs</author><pubDate>Tue, 17 Dec 2013 17:03:48 +0800</pubDate><description><![CDATA[#include 
#include 
#include 
#include 
using namespace std;
int cao[32768],pa[130000],stack[130000];
int p,ls;
void caonima(int v)
{
    v=cao[p]+v;
    if (v == -1)
        v=255;
    if (v == 256)
]]></description><category></category></item><item><title><![CDATA[【总结】2013年，再见ACM/ICPC]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/16910187</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/16910187</guid><author>sssogs</author><pubDate>Sat, 23 Nov 2013 23:27:48 +0800</pubDate><description><![CDATA[【总结】2013年，再见ACM/ICPC
plusplus7发表于 2013-11-23


在整理旧博客的文章时，偶然翻到了去年这个时候写的总结【总结】2012年，很水的一年，于是便想也写个今年的总结。

去年的那篇总结是写于去年打完ACM/ICPC亚洲区域赛成都赛区后。那时，我和Aikilis，小灯组队，队名叫3dmin/三弟马克思，满怀希望，以为能拿下学校的第一块亚洲区域赛奖牌.]]></description><category></category></item><item><title><![CDATA[【博客搬家】本博客已全面停止更新，新博客地址 plusplus7.com]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/16907287</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/16907287</guid><author>sssogs</author><pubDate>Sat, 23 Nov 2013 21:24:40 +0800</pubDate><description><![CDATA[新博客地址：http://www.plusplus7.com]]></description><category></category></item><item><title><![CDATA[川大校赛总结]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8877447</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8877447</guid><author>sssogs</author><pubDate>Thu, 02 May 2013 20:16:06 +0800</pubDate><description><![CDATA[对于这次四川大学的结果，我感到挺意外。


    比赛开始时，我们3人分工很明确。随后我很快读出了一道题，然后和tuber交流了题意，确认了以后就开始敲。此时Aikilis继续看题。随


后，很顺利地AC了此题。然后我们暂时没有发现简单的题，只找到一道BFS的题，但是实现起来比较复杂，于是决定让我敲此题，他们继续看


题。后来tuber读到了E题，发现是一道DP题，于是Ai]]></description><category></category></item><item><title><![CDATA[SGU 249 Matrix]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8836739</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8836739</guid><author>sssogs</author><pubDate>Mon, 22 Apr 2013 21:28:44 +0800</pubDate><description><![CDATA[格雷码。。#include 
#include 
#include 
#include 
using namespace std;
int grayCode[10000000];
int main()
{
    int s,n,m,pn,pm,t,i,j;
    scanf("%d%d",&n,&m);
    s=pow(2,(n+m));
    pn=pow(2,n);
    pm=p]]></description><category></category></item><item><title><![CDATA[LA 3618 Cubic Eight-Puzzle]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8836617</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8836617</guid><author>sssogs</author><pubDate>Mon, 22 Apr 2013 21:22:29 +0800</pubDate><description><![CDATA[立体的八数码问题。
        普通的DFS和BFS就别想了，迭代加深TLE，而用启发式搜索又想不到合适的启发函数。可发现目标状态有固定的2^8种，而状态转移的方式最多只有4种......所以用双向bfs，应该能解决问题。因为会产生很多重复的状态，所以用hash表判断一下重复的状态。stl的hash会TLE，所以还是手写静态链表哦。
#include 
#include 
#include]]></description><category></category></item><item><title><![CDATA[UVa 11212 Editing a Book]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8836606</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8836606</guid><author>sssogs</author><pubDate>Mon, 22 Apr 2013 21:03:16 +0800</pubDate><description><![CDATA[黑书P207 例题
IDA* 搜索，估价函数设计得很经典。
#include 
#include 
#include 
using namespace std;
class Board
{
public:
    int a[16];
    int h;
    int n;
    bool readIn()
    {
        int i;
        scanf("%d",]]></description><category></category></item><item><title><![CDATA[POJ 3856 deltree]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8836585</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8836585</guid><author>sssogs</author><pubDate>Mon, 22 Apr 2013 21:00:05 +0800</pubDate><description><![CDATA[模拟题。

deltree




Time Limit: 1000MS
 
Memory Limit: 65536K


Total Submissions: 58
 
Accepted: 33





Description
You have just run out of disk space and decided to delete some]]></description><category></category></item><item><title><![CDATA[POJ 2870 Light Up]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8836291</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8836291</guid><author>sssogs</author><pubDate>Mon, 22 Apr 2013 20:55:10 +0800</pubDate><description><![CDATA[题意简述：在n*m的格子中放置灯泡。其中格子中有些障碍，灯光可以照亮上下左右的格子，光线会被障碍挡住。
题目解法：搜索。
        每个状态的转移会产生很多状态，而目标状态有是不固定的，所以不能BFS或双向BFS。但是DFS又很容易陷入搜索到很深但是却又是错误的状态，而最优解的深度又是不深。所以考虑使用迭代加深搜索。
        剪枝：搜到每个障碍时，发现上面的格子是空的，剪枝。]]></description><category></category></item><item><title><![CDATA[POJ 3852 String LD]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8836273</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8836273</guid><author>sssogs</author><pubDate>Mon, 22 Apr 2013 19:54:36 +0800</pubDate><description><![CDATA[简单题，第一次提交居然TLE了。然后以为暴力不行，所以写了个Trie树，汗...赛后和别人讨论发现是可以暴力的。
#include 
#include 
#include 
using namespace  std;
int ans;
struct Trie
{
    int dep;
    int next[26];
    bool init(int d)
    {
        d]]></description><category></category></item><item><title><![CDATA[LA 5506 Eight]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8813386</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8813386</guid><author>sssogs</author><pubDate>Wed, 17 Apr 2013 13:07:28 +0800</pubDate><description><![CDATA[同样是South Central USA 1998的题，POJ和HDU上的数据比较水，用BFS不加任何优化都可以AC，时间一般是三位数左右。而用A*或IDA*则可以把时间控制在10ms左右，甚至0ms！
         然而UVa Live上的这道题是多数据的，用IDA*不加优化都会超时。除此之外题目输入输出很严格，case和case之间必须有空行，在最后一组case之后不能有空行，否则会Wr]]></description><category></category></item><item><title><![CDATA[LA 3211 Now or later]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8789472</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8789472</guid><author>sssogs</author><pubDate>Thu, 11 Apr 2013 17:34:28 +0800</pubDate><description><![CDATA[训练指南 P325 例题
2-SAT问题+二分。
题目的数据范围小，所以用朴素的算法解2-SAT能过。
tarjan的SCC算法判2-SAT
#include 
#include 
#include 
#include 
#define MAXN 2200
#define MAXN_2 4400
using namespace std;
int min(int a,int b)
{]]></description><category></category></item><item><title><![CDATA[UVa 11235 Frequent Values]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8789453</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8789453</guid><author>sssogs</author><pubDate>Thu, 11 Apr 2013 17:31:44 +0800</pubDate><description><![CDATA[训练指南 P198 例题
在一个非降序数组中寻找区间内出现次数最多的值的出现次数。
用线段树和RMQ都能AC。
RMQ解法
#include 
#include 
#include 
#define INF 0x7fffffff
using namespace std;
int n;
int a[120000];
int nb[120000],b[120000],bv[120000],l[]]></description><category></category></item><item><title><![CDATA[LA 3027 Corporative Network]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8789428</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8789428</guid><author>sssogs</author><pubDate>Thu, 11 Apr 2013 17:28:25 +0800</pubDate><description><![CDATA[并查集的应用。
在压缩路径的时候维护一个dist数组，保存其到祖先的长度。
#include 
#include 
#include 
using namespace std;
int uset[21000],dist[21000];
int abs(int a)
{
    return a>0?a:-a;
}
bool init()
{
    int i;
    for (i=0; i]]></description><category></category></item><item><title><![CDATA[LA 4329 Ping pong]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8789407</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8789407</guid><author>sssogs</author><pubDate>Thu, 11 Apr 2013 17:25:14 +0800</pubDate><description><![CDATA[训练指南 P197 例题
    扫描一下所有的i，计算出从1到i-1有多少个数比它大，和从i+1到n有多少个数比它大。
    然后就是交叉相乘算出总的方案数。
#include 
#include 
#include 
#define INF 120000
using namespace std;
long long a[21000];
long long l[21000],r[2100]]></description><category></category></item><item><title><![CDATA[LA 3942 Remember the Word]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8789386</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8789386</guid><author>sssogs</author><pubDate>Thu, 11 Apr 2013 17:21:50 +0800</pubDate><description><![CDATA[训练指南 P209 例题
用trie树使得每次寻找递推的下一步。
#include 
#include 
#include 
#define MOD 20071027
using namespace std;
struct Trie
{
    bool tag;
    int next[26];
    bool init()
    {
        tag=false;]]></description><category></category></item><item><title><![CDATA[POJ 3164 Command Network]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8744073</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8744073</guid><author>sssogs</author><pubDate>Sun, 31 Mar 2013 20:51:01 +0800</pubDate><description><![CDATA[朱刘算法模板题
#include 
#include 
#include 
#include 
#define eps 1e-8
#define INF 0x7fffffff
using namespace std;
struct point
{
    double x;
    double y;
};
struct edge
{
    int u;
    int v;
    doub]]></description><category></category></item><item><title><![CDATA[POJ 1442 Black Box]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8731028</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8731028</guid><author>sssogs</author><pubDate>Thu, 28 Mar 2013 13:38:06 +0800</pubDate><description><![CDATA[题意：给你n个数的数列，然后再给你m次查询，第i（i
题解：根据Discuss说的，这题可以用各种数据结构AC...为了练习一下红黑树，我写了个红黑树求解的。240行代码，结果居然1Y了,,,这...
解法：红黑树

//不想说什么了
#include 
#include 
#include 
using namespace std;
int m,n;
int a[60000]]]></description><category></category></item><item><title><![CDATA[POJ 1325 Machine Schedule]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8724502</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8724502</guid><author>sssogs</author><pubDate>Wed, 27 Mar 2013 00:33:14 +0800</pubDate><description><![CDATA[建图+Dinic=Accept
#include 
#include 
#include 
#include 
using namespace std;
int sta,end;
int val[300][300];
int tag[300];
int min(int a,int b)
{
    return a>b?b:a;
}
int bfs()
{
    int u,v;
    qu]]></description><category></category></item><item><title><![CDATA[CodeForces 1B Spreadsheet]]></title><link>https://blog.youkuaiyun.com/sssogs/article/details/8724493</link><guid>https://blog.youkuaiyun.com/sssogs/article/details/8724493</guid><author>sssogs</author><pubDate>Wed, 27 Mar 2013 00:31:20 +0800</pubDate><description><![CDATA[字符串处理+计算吧
用python，使用正则表达式来读取数据更方便一些。
import re
def convert(d):
    s=''
    while (d > 0):
        t=d%26
        d/=26
        if (t == 0):
            s+=('Z')
            d-=1
        else:]]></description><category></category></item></channel></rss>