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. 【MongoDB】MongoDB VS SQL数据库

    MongoDB和SQL数据库都能满足数据库的基本功能:1.有组织的存放数据:2.按照需求查询数据 传统的SQL数据库(e.g.Oracle, MySQL) 对表的运用不够灵活,横向扩展不太容易,而它的 ...

  2. Hadoop2.0重启脚本

    Hadoop2.0重启脚本 方便重启带ha的集群,写了这个脚本 #/bin/bash sh /opt/zookeeper-3.4.5-cdh4.4.0/bin/zkServer.sh restart ...

  3. spring--事务原理

    Spring支持以下7种事务传播行为. 传播行为 XML文件 propagation值 含义 PROPAGATION_REQUIRED REQUIRED 表示当前方法必须在一个具有事务的上下文中运行. ...

  4. POJ 2516 Minimum Cost 最小费用流

    题目: 给出n*kk的矩阵,格子a[i][k]表示第i个客户需要第k种货物a[i][k]单位. 给出m*kk的矩阵,格子b[j][k]表示第j个供应商可以提供第k种货物b[j][k]单位. 再给出k个 ...

  5. 初识 Asp.Net内置对象之Application对象

    Application对象 Applocation对象用于共享应用程序级信息,即多个用户可以共享一个Applocation对象. 用户在请求Asp.Net文件时,将启动应用程序并且创建Applicat ...

  6. 上架第一个APP到苹果商店被拒绝5次

    - : Metadata Rejected (APP中的注册时跳转的 - 用户协议视图没有内容).Waiting For Review 6天  In Review 1天 第二次被拒绝 -- : Met ...

  7. 304 CORS

    304响应, CORS问题: 没有 Access-Control-Allow-Origin 这个头信息时,以前次返回的200请求为准. 示例:可能已被删除 http://7af3zm.com1.z0. ...

  8. JavaScript--Function类型(11)

    // 在JS中,Function(函数)类型实际上是对象;每个函数都是Function类型的实例;而且都与其他引用类型一样具有属性和方法; // 由于函数是对象,因此函数名实际上也是一个指向函数对象的 ...

  9. Part 8 AngularJS filters

    Filters in angular can do 3 different things 1. Format data 2. Sort data 3. Filter data Filters can ...

  10. JQuery 动态添加onclick事件

    $('#div_id').click(function(){ show(1,2,this); });