HDOJ 4642 Fliping game

Fliping game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1770    Accepted Submission(s): 1157

Problem Description

Alice and Bob are playing a kind of special game on an N*M board (N rows, M columns). At the beginning, there are N*M coins in this board with one in each grid and every coin may be upward or downward freely. Then they take turns to choose a rectangle (x₁, y₁)-(n, m) (1 ≤ x₁≤n, 1≤y₁≤m) and flips all the coins (upward to downward, downward to upward) in it (i.e. flip all positions (x, y) where x₁≤x≤n, y₁≤y≤m)). The only restriction is that the top-left corner (i.e. (x₁, y₁)) must be changing from upward to downward. The game ends when all coins are downward, and the one who cannot play in his (her) turns loses the game. Here's the problem: Who will win the game if both use the best strategy? You can assume that Alice always goes first.

 

Input

The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case starts with two integers N and M indicate the size of the board. Then goes N line, each line with M integers shows the state of each coin, 1<=N,M<=100. 0 means that this coin is downward in the initial, 1 means that this coin is upward in the initial.

 

Output

For each case, output the winner’s name, either Alice or Bob.

 

Sample Input

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

 

Sample Output

Alice Bob

 

題意就是一局遊戲，在n*m的棋盤裏面有硬幣，分爲朝上或者朝下，每個人可以選擇一個x,y，翻轉x到n，y到m的硬幣,求最後誰能獲勝。

很少做博弈論，其實這個題很簡單，每次翻轉時都會把n，m的硬幣翻轉，要使最後nm==1；

如果輸入nm爲0那麼就肯定需要奇數次，反之就需要偶數次。(我還是太年輕)

#include<stdio.h>
#include<algorithm>
#include<string>
#include<string.h>
#include<queue>
#include<stack>
#include<math.h>
using namespace std;
int main()
{
    int a[105][1005];
    int n,m,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
                scanf("%d",&a[i][j]);
        }
        if(a[n][m]==1)
            printf("Alice\n");
        else
            printf("Bob\n");
    }
return 0;
}