A Chess Game
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 3791   Accepted: 1549

Description

Let's design a new chess game. There are N positions to hold M chesses in this game. Multiple chesses can be located in the same position. The positions are constituted as a topological graph, i.e. there are directed edges connecting some positions, and no cycle exists. Two players you and I move chesses alternately. In each turn the player should move only one chess from the current position to one of its out-positions along an edge. The game does not end, until one of the players cannot move chess any more. If you cannot move any chess in your turn, you lose. Otherwise, if the misfortune falls on me... I will disturb the chesses and play it again.

Do you want to challenge me? Just write your program to show your qualification!

Input

Input contains multiple test cases. Each test case starts with a number N (1 <= N <= 1000) in one line. Then the following N lines describe the out-positions of each position. Each line starts with an integer Xi that is the number of out-positions for the position i. Then Xi integers following specify the out-positions. Positions are indexed from 0 to N-1. Then multiple queries follow. Each query occupies only one line. The line starts with a number M (1 <= M <= 10), and then come M integers, which are the initial positions of chesses. A line with number 0 ends the test case.

Output

There is one line for each query, which contains a string "WIN" or "LOSE". "WIN" means that the player taking the first turn can win the game according to a clever strategy; otherwise "LOSE" should be printed.

Sample Input

4
2 1 2
0
1 3
0
1 0
2 0 2
0 4
1 1
1 2
0
0
2 0 1
2 1 1
3 0 1 3
0

Sample Output

WIN
WIN
WIN
LOSE
WIN
/*
poj 2425 AChessGame(博弈) 给你一个有向的图,上面的棋子可以移动到下一个节点,如果当前无法移动则失败 可以同dfs求出所有节点的sg值,然后进行计算即可 hhh-2016-08-02 16:50:29 4
2 1 2
0
1 3
0
1 0
2 0 2
0 4
1 1
1 2
0
0
2 0 1
2 1 1
3 0 1 3
0 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <functional>
typedef long long ll;
#define lson (i<<1)
#define rson ((i<<1)|1)
using namespace std; const int maxn = 1000+10; int sg[maxn];
int Map[maxn][maxn];
int n;
void dfs(int now)
{
int vis[maxn] = {0};
for(int i = 0; i < n; i++)
{
if(Map[now][i])
{
if(sg[i] == -1)
dfs(i);
vis[sg[i]] = 1;
} }
for(int i = 0; i < n; i++)
{
if(!vis[i])
{
sg[now] = i;
break;
}
}
} int main()
{
int x,m;
while(scanf("%d",&n) != EOF && n)
{
memset(sg,-1,sizeof(sg));
memset(Map,0,sizeof(Map));
for(int i = 0; i < n; i++)
{
scanf("%d",&m);
for(int j = 1; j <= m; j++)
{
scanf("%d",&x);
Map[i][x] = 1;
}
}
for(int i = 0; i < n; i++)
{
if(sg[i] == -1)
dfs(i);
}
while(scanf("%d",&m)!=EOF && m)
{
int ans = 0;
for(int i = 0; i < m; i++)
{
scanf("%d",&x);
ans ^= sg[x];
}
if(ans)
printf("WIN\n");
else
printf("LOSE\n");
}
}
return 0;
}

  

poj 2425 AChessGame(博弈)的更多相关文章

  1. poj 1704 阶梯博弈

    转自http://blog.sina.com.cn/s/blog_63e4cf2f0100tq4i.html 今天在POJ做了一道博弈题..进而了解到了阶梯博弈...下面阐述一下我对于阶梯博弈的理解. ...

  2. POJ 2425 A Chess Game#树形SG

    http://poj.org/problem?id=2425 #include<iostream> #include<cstdio> #include<cstring&g ...

  3. POJ 2425 A Chess Game 博弈论 sg函数

    http://poj.org/problem?id=2425 典型的sg函数,建图搜sg函数预处理之后直接求每次游戏的异或和.仍然是因为看不懂题目卡了好久. 这道题大概有两个坑, 1.是搜索的时候vi ...

  4. POJ 2960 S-Nim<博弈>

    链接:http://poj.org/problem?id=2960 #include<stdio.h> #include<string.h> ; ; int SG[N];//S ...

  5. poj 2425 A Chess Game_sg函数

    题意:给你一个有向无环图,再给你图上的棋子,每人每次只能移动一个棋子,当轮到你不能移动棋子是就输了,棋子可以同时在一个点 比赛时就差这题没ak,做了几天博弈终于搞懂了. #include <io ...

  6. POJ 2234 Nim博弈

    思路: nim博弈裸题 xor一下 //By SiriusRen #include <cstdio> using namespace std; int n,tmp,xx; int main ...

  7. poj 2425 A Chess Game 博弈论

    思路:SG函数应用!! 代码如下: #include<iostream> #include<cstdio> #include<cmath> #include< ...

  8. [原博客] POJ 2425 A Chess Game

    题目链接题意:给定一个有向无环图(DAG),上面放有一些旗子,旗子可以重合,两个人轮流操作,每次可以把一个旗子从一个位置移动到相邻的位置,无法移动时输,询问先手是否必胜. 这道题可以把每个旗子看作单独 ...

  9. poj 2425 A Chess Game(SG函数)

    A Chess Game Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 3551   Accepted: 1440 Desc ...

随机推荐

  1. networkx 学习

    import networkx as nx import pylab import numpy as np #自定义网络 row=np.array([,,,,,,]) col=np.array([,, ...

  2. 【Swift】Runtime动态性分析

    Swift是苹果2014年发布的编程开发语言,可与Objective-C共同运行于Mac OS和iOS平台,用于搭建基于苹果平台的应用程序.Swift已经开源,目前最新版本为2.2.我们知道Objec ...

  3. Flask 学习 六 大型程序结构

    pip freeze >requirement.txt 自动生成版本号 pip install -r requirement.txt 自动下载对应的库 梳理结构 config.py #!/usr ...

  4. java 二维码解析和生成

    package ykxw.web.qrcode.utils; import java.awt.Color; import java.awt.Graphics2D; import java.awt.im ...

  5. js实现短暂提示框

    业务场景:当鼠标移入某元素时,显示提示框进行介绍.当鼠标移除时,会自动消失.引入ToolTip.js和ToolTip.css 主方法:ToolTip.show(需要提示的元素id, 随意不重复即可, ...

  6. C# reportview 按时间改变行颜色

    //) AND ((Day(Now()) - Day() AND (Day(Now()) - Day()),) AND (Day(Now()) - Day()) OR (Month(Now()) - ...

  7. Centos6.7的在虚拟机virulBox下的lamp平台的搭建

    实验环境: linux:小甲鱼带你学C语言,带你飞的提供的体积比较小的centos6.7和virtualBox mysql,apahce,php是燕十八在Linux基础进阶中提供的安装方式: 结果,安 ...

  8. Class-Based-View(CBV)

    我们都知道,Python是一个面向对象的编程语言,如果只用函数来开发,有很多面向对象的优点就错失了(继承.封装.多态).所以Django在后来加入了Class-Based-View.可以让我们用类写V ...

  9. GIT的安装及命令使用

    http://blog.jobbole.com/78960/ 因此:多人协作工作模式一般是这样的: 首先,可以试图用git push origin branch-name推送自己的修改. 如果推送失败 ...

  10. GIT入门笔记(18)- 标签创建和管理

    git tag <name>用于新建一个标签,默认为HEAD,也可以指定一个commit id: git tag -a <tagname> -m "blablabla ...