标签：cstring amp integer blog style 不能 rip des 状态
从这开始我们来进入做题环节！作为一个较为抽象的知识点，博弈论一定要结合题目才更显魅力。今天，我主要介绍一些经典的题目，重点是去理解模型的转化，sg函数的推理和证明。话不多说，现在开始！
Georgia and Bob
| Time Limit: 1000MS |
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Memory Limit: 10000K |
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Description
Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2, 3, ..., and place N chessmen on different grids, as shown in the following figure for example:
Georgia and Bob move the chessmen in turn. Every time a player will choose a chessman, and move it to the left without going over any other chessmen or across the left edge. The player can freely choose number of steps the chessman moves, with the constraint that the chessman must be moved at least ONE step and one grid can at most contains ONE single chessman. The player who cannot make a move loses the game.
Georgia always plays first since "Lady first". Suppose that Georgia and Bob both do their best in the game, i.e., if one of them knows a way to win the game, he or she will be able to carry it out.
Given the initial positions of the n chessmen, can you predict who will finally win the game?
Input
The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case contains two lines. The first line consists of one integer N (1 <= N <= 1000), indicating the number of chessmen. The second line contains N different integers P1, P2 ... Pn (1 <= Pi <= 10000), which are the initial positions of the n chessmen.
Output
For each test case, prints a single line, "Georgia will win", if Georgia will win the game; "Bob will win", if Bob will win the game; otherwise ‘Not sure‘.
Sample Input
2
3
1 2 3
8
1 5 6 7 9 12 14 17
Sample Output
Bob will win
Georgia will win
大致题意是说给一个很长的棋盘，一些地方有棋子，每个格子只能放1个棋子。每次必须要向左移动1个棋子，但不能移除棋盘，也不能超过它左边的第一个棋子。求先手是否必胜。
题解：
(检查草稿箱突然发现自己有暑假的博客没发出来,尴尬......)
这道题上来硬想肯定什么都想不出来。
我们只能通过由浅入深的推理才能做出这道题。
首先我们考虑必败状态的定义：
对于某两个棋子，如果他们两个靠在了一起，那么它们对应的状态就是一个必败状态。
这一点很显然，如果两个棋子贴在一起，先手只能移动前面的棋子，而后手可以通过紧跟先手来继续使先手拿到必败状态。
那么这样，命题得证，这两个紧贴的石子就是一个必败状态了。
那么我们考虑它是从哪里转移而来的：两个棋子如果没有距离了，那么它肯定是从一开始有距离的游戏状态转移过来的.
那么我们可以得到一些式子：
sg(距离为1)=mex(sg(距离为0))=1,
sg(距离为2)=mex(sg(距离为0),sg(距离为1))=2,
sg(距离为3)=mex(sg(距离为0),sg(距离为1),sg(距离为2))=3.......
到这里，读者应该想到了什么了：这就是一个nim游戏的变种！
因此，我们把2个棋子看做1组，之间的空位数看做一堆石子，最后按照nim游戏计算即可。代码见下：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5 const int N=1010;
6 int a[N];
7 inline bool mt(const int &a,const int &b){return a<b;}
8 int main()
9 {
10 int t,n;scanf("%d",&t);
11 while(t--)
12 {
13 scanf("%d",&n);int ans=0;
14 for(int i=1;i<=n;i++)
15 scanf("%d",&a[i]);
16 sort(a+1,a+n+1,mt);
17 for(int i=1;i<=n;i++)
18 if( ((i&1)&&(n&1)) || (!(i&1)&&!(n&1)) )ans^=a[i]-a[i-1]-1;
19 if(!ans)printf("Bob will win\n");
20 else printf("Georgia will win\n");
21 }
22
23 }
[POJ1704]Georgia and Bob 博弈论
标签：cstring amp integer blog style 不能 rip des 状态
原文地址：http://www.cnblogs.com/LadyLex/p/7260629.html
