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
- HTML中的Head标签学习
在页面加载完成的时候,标签head里的内容,是不会在页面中显示出来的.它包含了像页面的<title>(标题) ,CSS(如果你想用CSS来美化页面内容),图标和其他的元数据(比如 作者,关 ...
- How To Install Cacti On Ubuntu 14
How To Install Cacti On Ubuntu 14.04/14.10 by anismaj Cacti is an open source web based network moni ...
- 通过Application存取公共数据比如登录信息等..
Android系统在运行每一个程序应用的时候,都会创建一个Application对象,用于存储与整个应用相关的公共变量.一个Android应用只会生成一个Application对象,在不同的Activ ...
- Android 线程池的类型、区别以及为何要用线程池
每个 Android 应用进程在创建时,会同时创建一个线程,我们称之为主线程,负责更新 UI 界面以及和处理用户之间的交互,因此,在 Android 中,我们又称之为 UI 线程.一个进程中 UI 线 ...
- [翻译] DZNSegmentedControl
DZNSegmentedControl A drop-in replacement for UISegmentedControl for showing counts, to be used typi ...
- CSV输入输出
读取csv文件: import csv cf = open('D:\pywe.csv','rb') cf.readline() #读取标题行,光标移动到下一行(相当于调过标题行) for l in c ...
- Python学习---装饰器的学习1210
装饰器的基础 学习前提: 作用域 + 函数的理解 + 闭包 [学习,理解] 代码编写原则: 对修改开放对扩展开放 装饰器本质上是一个函数,该函数用来处理其他函数,它可以让其他函数在不需要修改代码的前 ...
- php图片处理类
代码 Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/-- ...
- ZT pthread_detach
pthread_detach 创 建一个线程默认的状态是joinable, 如果一个线程结束运行但没有被join,则它的状态类似于进程中的Zombie Process,即还有一部分资源没有被回收(退出 ...
- Python 处理脚本的命令行参数-getopt
# -*- coding:utf-8 -*- import sys def test(): """ 参数列表:sys.argv 参数个数:len(sys.argv) 脚本 ...