链接:http://acm.hdu.edu.cn/showproblem.php?pid=5724

题意:一个n*20的棋盘,n <= 1000;棋盘上有一些棋子,每颗棋子只能移动到右边的第一个空格。不能移动者输;其中 Alice先手;如果Alice能赢输出"YES";

思路:每个子游戏的大小只有20,使用状压即可;但是每个子游戏的SG值需要建立在比其规模更小的SG值之上;这样,在二进制高位变为0低位变为1之后,显然数值变大了;

那么就二进制低位表示高位编码即可;

预处理出所有的有效状态的SG值之后,使用NIM和看ans 是否为0即可;为0表示先手输;

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define inf 0x3f3f3f3f
#define A first
#define B second
#define MK make_pair
#define esp 1e-8
#define zero(x) (((x)>0?(x):-(x))<eps)
#define bitnum(a) __builtin_popcount(a)
#define clear0 (0xFFFFFFFE)
#define mod 1000000007
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
template<typename T>
void read1(T &m)
{
T x = 0,f = 1;char ch = getchar();
while(ch <'0' || ch >'9'){ if(ch == '-') f = -1;ch=getchar(); }
while(ch >= '0' && ch <= '9'){ x = x*10 + ch - '0';ch = getchar(); }
m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
if(a>9) out(a/10);
putchar(a%10+'0');
}
inline ll gcd(ll a,ll b){ return b == 0? a: gcd(b,a%b); }
int SG[1<<20], state[22];
void init()
{
for(int i = 0; i < 1<<20; i++){
MS0(state);
rep_1(j,19,0){
if(i & (1<<j)){
rep_1(k,j-1,0) if(!(i & (1 << k))){
state[SG[i^(1<<k)^(1<<j)]] = 1;
break;
}
}
}
rep1(j,0,19) if(state[j] == 0){
SG[i] = j;break;
}
}
}
int main()
{
//freopen("data.txt","r",stdin);
//freopen("out.txt","w",stdout);
init();
int T, kase = 1;
scanf("%d",&T);
while(T--){
int n, m, p, ans = 0;
read1(n);
rep1(i,1,n){
read1(m);
int S = 0;
while(m--){
read1(p);
S |= 1 << 20 - p;
}
ans ^= SG[S];
}
puts(ans?"YES":"NO");
}
return 0;
}

2016 Multi-University Training Contest 1 Chess 组合游戏+状压(预处理)的更多相关文章

  1. 2016 Al-Baath University Training Camp Contest-1

    2016 Al-Baath University Training Camp Contest-1 A题:http://codeforces.com/gym/101028/problem/A 题意:比赛 ...

  2. 2016 Al-Baath University Training Camp Contest-1 E

    Description ACM-SCPC-2017 is approaching every university is trying to do its best in order to be th ...

  3. 2016 Al-Baath University Training Camp Contest-1 A

    Description Tourist likes competitive programming and he has his own Codeforces account. He particip ...

  4. 2016 Al-Baath University Training Camp Contest-1 J

    Description X is fighting beasts in the forest, in order to have a better chance to survive he's gon ...

  5. 2016 Al-Baath University Training Camp Contest-1 I

    Description It is raining again! Youssef really forgot that there is a chance of rain in March, so h ...

  6. 2016 Al-Baath University Training Camp Contest-1 H

     Description You've possibly heard about 'The Endless River'. However, if not, we are introducing it ...

  7. 2016 Al-Baath University Training Camp Contest-1 G

    Description The forces of evil are about to disappear since our hero is now on top on the tower of e ...

  8. 2016 Al-Baath University Training Camp Contest-1 F

    Description Zaid has two words, a of length between 4 and 1000 and b of length 4 exactly. The word a ...

  9. 2016 Al-Baath University Training Camp Contest-1 D

    Description X is well known artist, no one knows the secrete behind the beautiful paintings of X exc ...

随机推荐

  1. 补充:tableView优化总结

    tableView优化方式总结 (说明:一下只是我自己的小总结,会有不足,望大神指正,多谢大家了) 1 缓存 cell 的行高.目的方便第二遍访问 cell. 2 减少图片的blend 操作.包括:设 ...

  2. Pull解析-解析xml文件

    首先需要导入jar包:kxml2-2.2.2.jar 例程: main: /** * pull解析 * * @author my * */ public class DemoParserStudent ...

  3. spring工作机制

    Hibernate.struts,还差一个spring 就一起发出去.. spring工作机制及为什么要用? 1.springmvc请所有的请求都提交给DispatcherServlet,它会委托应用 ...

  4. Oracle基础(九) Oracle的体系结构

    一.Oracle体系结构概述: Oracle的体系结构是指数据库的组成.工作过程与原理,以及数据在数据库中的组织与管理机制.要了解Oracle数据库的体系结构,必须理解Oracle系统的重要概念和主要 ...

  5. JavaScript 编写线程代码引用Concurrent.Thread.js

    马上来下载和使用源码吧!假定你已经将下载的源码保存到一个名为Concurrent.Thread.js的文件夹里,在进行任何操作之前,先运行如下程序,这是一个很简单的功能实现: <script t ...

  6. PHP使用外部命令导出数据库,备份到服务器并下载到本地

    <?php // $dumpFileName目录要有可写权限 $DbHost = 'localhost'; $DbUser = 'root'; $DbPwd = '123456'; $DbNam ...

  7. oracle-SQL语言基础-事务控制命令命令

    事务控制命令命令 COMMITROLLBACKSAVEPOINTSET TRANSACTION 当第一条可执行的SQL语句开始执行,数据库事务就开始.随着下面任一事件发生,数据库事务结束:执行COMM ...

  8. oracle-同义词Synonyms + 用户访问控制(grant 和 revoke)

    同义词(Synonyms) 创建同义词:    语法    CREATE [PUBLIC] SYNONYM synonym        FOR    object; CREATE SYNONYM   ...

  9. Unity3d之按键

    if (Input.GetKeyDown(KeyCode.A)){ Debug.Log("您按下了A键"); } if (Input.GetKeyUp(KeyCode.A)) { ...

  10. js打印Iframe中的内容,并且不需要预览。

    js打印Iframe中的内容,并且不需要预览 js代码如下: <script type="text/javascript" language="Javascript ...