POJ2311 Cutting Game(博弈论)
总时间限制: 1000ms 内存限制: 65536kB
描述
Urej loves to play various types of dull games. He usually asks other people to play with him. He says that playing those games can show his extraordinary wit. Recently Urej takes a great interest in a new game, and Erif Nezorf becomes the victim. To get away from suffering playing such a dull game, Erif Nezorf requests your help. The game uses a rectangular paper that consists of W*H grids. Two players cut the paper into two pieces of rectangular sections in turn. In each turn the player can cut either horizontally or vertically, keeping every grids unbroken. After N turns the paper will be broken into N+1 pieces, and in the later turn the players can choose any piece to cut. If one player cuts out a piece of paper with a single grid, he wins the game. If these two people are both quite clear, you should write a problem to tell whether the one who cut first can win or not.
输入
The input contains multiple test cases. Each test case contains only two integers W and H (2 <= W, H <= 200) in one line, which are the width and height of the original paper.
输出
For each test case, only one line should be printed. If the one who cut first can win the game, print "WIN", otherwise, print "LOSE".
样例输入
2 2
3 2
4 2
样例输出
LOSE
LOSE
WIN
来源
POJ Monthly,CHEN Shixi(xreborner)
题解:
用sg函数做,还套了个记忆化
切成两半以后,返回的sg值就是两个数的异或。如果有两个必胜,用掉一个,别人再用一个,你就是两个必败了
注意点:为什么要从2开始枚举?
因为如果是1了,先手必胜(非(1,1),可是我的初始化没有设置这种情况,就会出现-1
#include <bits/stdc++.h>
#define int long long
using namespace std;
int sg[230][230], x, y, tong[1200];
bool vis[230][230];
int find(int x, int y) {
if(sg[x][y]!=-1) return sg[x][y];
memset(tong, 0, sizeof tong);
for (int i = 2; i <= x-i; i++)
tong[(find(i, y)) ^ (find(x - i, y))]=1;
for (int i = 2; i <= y-i; i++)
tong[(find(x, i)) ^ (find(x, y - i))]=1;
for (int i = 0; i < 1200; i++)
if (!tong[i]) {
sg[x][y] = i;
return i;
}
}
signed main() {
memset(sg,-1,sizeof sg);
sg[2][2] = sg[2][3] = sg[3][2] = 0;
while (scanf("%lld%lld", &x, &y) != EOF) {
if (!find(x, y))
puts("LOSE");
else
puts("WIN");
}
return 0;
}
POJ2311 Cutting Game(博弈论)的更多相关文章
- [poj2311]Cutting Game_博弈论
Cutting Game poj-2311 题目大意:题目链接 注释:略. 想法: 我们发现一次操作就是将这个ICG对应游戏图上的一枚棋子变成两枚. 又因为SG定理的存在,记忆化搜索即可. 最后,附上 ...
- $POJ2311\ Cutting\ Game$ 博弈论
正解:博弈论 解题报告: 传送门! 首先看到说,谁先$balabala$,因为$SG$函数是无法解决这类问题的,于是考虑转化成"不能操作者赢/输"的问题,不难想到先剪出$1\cdo ...
- 【博弈论】【SG函数】poj2311 Cutting Game
由于异或运算满足结合律,我们把当前状态的SG函数定义为 它所能切割成的所有纸片对的两两异或和之外的最小非负整数. #include<cstdio> #include<set> ...
- POJ2311 Cutting Game 博弈 SG函数
Cutting Game Description Urej loves to play various types of dull games. He usually asks other peopl ...
- POJ2311 Cutting Game
题意 Language:Default Cutting Game Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6007 Acc ...
- poj 2311 Cutting Game 博弈论
思路:求SG函数!! 代码如下: #include<iostream> #include<cstdio> #include<cmath> #include<c ...
- 博弈论BOSS
基础博弈的小结:http://blog.csdn.net/acm_cxlove/article/details/7854530 经典翻硬币游戏小结:http://blog.csdn.net/acm_c ...
- 博弈问题之SG函数博弈小结
SG函数: 给定一个有向无环图和一个起始顶点上的一枚棋子,两名选手交替的将这枚棋子沿有向边进行移动,无法移 动者判负.事实上,这个游戏可以认为是所有Impartial Combinatorial Ga ...
- 【Mark】博弈类题目小结(HDU,POJ,ZOJ)
转载请注明出处,谢谢http://blog.csdn.net/ACM_cxlove?viewmode=contents by---cxlove 首先当然要献上一些非常好的学习资料: 基础博弈的小 ...
随机推荐
- ContentProvider与ContentResolver
使用ContentProvider共享数据: 当应用继承ContentProvider类,并重写该类用于提供数据和存储数据的方法,就可以向其他应用共享其数据.虽然使用其他方法也可以对外共享数据,但数据 ...
- 一张图看懂 JS 原型链
JS 原型链,画了张图,终于理清楚各种关系有木有 写在最后: __proto__是每个对象都有的一个属性,而prototype是函数才会有的属性!!! function Person() { } 是函 ...
- sqlserver 查询表结构及字段说明
SELECT [Table Name] = OBJECT_NAME(c.object_id), [Column Name] = c.name, [Description] = ex.value FRO ...
- LDF文件丢失, 如何仅用MDF文件恢复数据库呢?
笔者的一个大小为2 TB的SQL Server的database的LDF文件在玩存储盘映射的过程中莫名其妙的丢失了. 好在MDF文件还在. 笔者慌了, Bruce Ye告诉笔者, 不用着急, 光用MD ...
- 深入浅出SharePoint2012——安装Report Service
安装顺序 Microsoft .NET Framework 3.5 SP1 report service installation,pls SQLServer2008R2SP1-KB2528583-x ...
- [EffectiveC++]导读 default构造函数
class B { public: explicit B(int x = 0,bool b = true); //default构造函数 }; explicit可以阻止用来执行隐式类型转换,但是可以用 ...
- [原]外网访问用azure虚拟机搭建的网站
1.Ubuntu+Apache+PHP的环境搭建(此处省去了mysql的步骤) 在azure上搭建上述的开发环境和在本地PC搭建是一样的步骤,具体介绍请参看这里. 2.从外网访问 注意,这一步的前提是 ...
- 导出当前域内所有用户hash的技术整理
0x00目标: 导出当前域内所有用户的hash 0x01测试环境: 域控:server2008 r2 杀毒软件:已安装* 域控权限:可使用net use远程登陆,不使用3389 0x02测试方法: ( ...
- github air项目中遇到的几个问题及解决(nodejs居多)
https://github.com/cambecc/air 1.按照github中给出的步骤,执行到npm install,项目中的package.json包含了要安装的包的版本,但是安装的时候,p ...
- [XML123] FpML
Wiki Fpml FpML (Financial products Markup Language) is a business information exchange standard base ...