Georgia and Bob POJ - 1704 （变形的nim博弈）

最新推荐文章于 2022-02-02 21:10:30 发布

elbadaernu

最新推荐文章于 2022-02-02 21:10:30 发布

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本文介绍了一种两人交替移动棋子的游戏，通过分析棋子初始位置判断胜负。采用nim博弈思想，将棋子两两配对，计算每对棋子间的距离并进行异或操作，最终结果用于预测胜者。

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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:
这里写图片描述
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：从左到右有n个棋子，告诉你每个棋子的位置。每个棋子只能向左移动，而且不能越过在它左边的棋子，当棋子移动到最左边1号位置时停止。现在两个人轮流选择一个棋子进行移动，谁先无法移动棋子谁输，Georgia先手，问输赢情况。

思路：如果棋子数为偶数，那么我们可以将其两两抱团来考虑，假设a移动了一个棋子，那么b可以移动和它抱团的棋子相同的距离，所以每个团和团之间的距离对结果就不会产生影响了，所以只需考虑每对棋子之间的距离，转化成nim博弈。如果棋子数量是奇数个的话，就增加一个1号位的棋子（对结果无影响）然后同理。

代码如下

#include <iostream> 
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
#include <cstdio>
#include <map>
using namespace std; 
#define ll long long int 

int wei[1005];
int main()  
{    
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)
        cin>>wei[i];

        sort(wei+1,wei+1+n);

        int sum=0;
        int i;
        wei[0]=0;
        for(i=n;i>0;i-=2)
        {
            sum^=(wei[i]-wei[i-1]-1);
        }
        if(sum)
        cout<<"Georgia will win\n";
        else
        cout<<"Bob will win\n";
    }
}