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
- Java中的数据类型转换
先来看一个题: Java类Demo中存在方法func0.func1.func2.func3和func4,请问该方法中,哪些是不合法的定义?( ) public class Demo{ float fu ...
- C++学习笔记(7)----类的数组中构造函数和析构函数的调用顺序
C++类的数组中构造函数和析构函数的调用顺序(2) 对于如下的代码: #include<iostream> using namespace std; class CBase { priva ...
- Python - Exceptions
官方文档:https://docs.python.org/3/library/exceptions.html 1. 使用try...except... 2. 输出错误信息的方式为: try: curs ...
- CentOS7安装minio
[root@localhost ~]# wget https://dl.minio.io/server/minio/release/linux-amd64/minio -bash: wget: 未找到 ...
- android头像上传(获取头像加剪切)
因为项目中需要用到头像上传的功能,所以就下个Ddmo先来实现下. demo我是类似仿微信的,在一个GridView中展示所有的图片,其中第一个item可以去照相:获取到图片后再进行剪切. 图片的剪切是 ...
- 平均负载(Load average)
load average 的含义平均负载(load average)是指系统的运行队列的平均利用率,也可以认为是可运行进程的平均数. top命令中load average显示的是最近1分钟.5分钟和1 ...
- 自动化测试基础篇--Selenium框架设计(POM)
一.自动化测试框架 感谢木棉花的漂泊分享,内容转自链接:http://www.cnblogs.com/fengyiru6369/p/8053035.html 1.什么是自动化测试框架 简单来说,自动化 ...
- Django 的路由层URL 分组 路由分发 反向解析
URL配置(URLconf)就像Django 所支撑网站的目录.它的本质是URL与要为该URL调用的视图函数之间的映射表:你就是以这种方式告诉Django,对于这个URL调用这段代码,对于那个URL调 ...
- 如何理解springaop
初看aop,上来就是一大堆术语,而且还有个拉风的名字,面向切面编程,都说是oop的一种有益补充等等,一下子让你不知所措,心想着:怪不得很多人都和我说aop多难多难.当我看进去以后,我才发现:它就是一些 ...
- MYSQL数据库、用户、表等基础构建
MYSQL数据库.用户.表等基础构建: 1.->:创建数据库: 1.1. create schema [数据库名称] default character set utf8 collate utf ...