Yet Another Game of Stones ZOJ - 3964 (博弈)

Alice and Bob are playing yet another game of stones. The rules of this game are as follow:

The game starts with n piles of stones indexed from 1 to n. The i-th pile contains ai stones and a special constraint indicated as bi.
The players make their moves alternatively. The allowable moves for the two players are different.
An allowable move of Bob is considered as removal of some positive number of stones from a pile.
An allowable move of Alice is also considered as removal of some positive number of stones from a pile, but is limited by the constraint bi of that pile.
If bi = 0, there are no constraints.
If bi = 1, Alice can only remove some odd number of stones from that pile.
If bi = 2, Alice can only remove some even number of stones from that pile.
Please note that there are no constraints on Bob.
The player who is unable to make an allowable move loses.
Alice is always the first to make a move. Do you know who will win the game if they both play optimally?

Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1 ≤ n ≤ 105), indicating the number of piles.

The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 109), indicating the number of stones in each pile.

The third line of each test case contains n integers b1, b2, …, bn (0 ≤ bi ≤ 2), indicating the special constraint of each pile.

It is guaranteed that the sum of n over all test cases does not exceed 106.

We kindly remind you that this problem contains large I/O file, so it’s recommended to use a faster I/O method. For example, you can use scanf/printf instead of cin/cout in C++.

Output
For each test case, output “Alice” (without the quotes) if Alice will win the game. Otherwise, output “Bob” (without the quotes).

Sample Input
3
2
4 1
1 0
1
3
2
1
1
2
Sample Output
Alice
Bob
Bob
**思路：**nim博弈的变形问题。
代码:

#include<iostream>
#include<cstdio>
#include<stack>
#include<algorithm>
#include<queue>
#include<cstring>
#include<cmath>

using namespace std;
int a[100005];
int b[100005];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int countt=0;
        bool sign=false;
        int x;
        int n;
        cin>>n;
        for(int i=0;i<n;i++)
            scanf("%d",a+i);
        for(int i=0;i<n;i++)
            scanf("%d",b+i);
        for(int i=0;i<n;i++)
        {
            if(a[i]%2&&b[i]==2)
                sign=true;
            if((a[i]>1&&b[i]==1)||(a[i]%2==0&&b[i]==2))
            {
                countt++;
                if(a[i]%2==0&&b[i]==1)
                    a[i]=1;
                else
                    a[i]=0;
            }
        }
        if(sign||countt>1)
        {
            cout<<"Bob"<<endl;
            continue;
        }
        int ans=0;
        for(int i=0;i<n;i++)
            ans^=a[i];
        if(countt==0)
            cout<<(ans?"Alice":"Bob")<<endl;
        else
            cout<<(ans==0?"Alice":"Bob")<<endl;
    }
    return 0;
}