hdu2873 Bomb Game-----sg

Bomb Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 353 Accepted Submission(s): 165

Problem Description
John and Jack, two mathematicians, created a game called “Bomb Game” at spared time. This game is played on an n*m chessboard. A pair of integers (p, q) represents the grid at row p, column q. Some bombs were placed on the chessboard at the beginning. Every round, a player can choose to explode a bomb located at (p, q), and the exploded bomb will disappear. Furthermore:

1.If p>1 and q>1, the bomb will split up into two bombs located at (u, q) and (p, v), u<p, v<q, u and v are chosen by the player.
2.If p=1 and q>1, one new bomb will be produced, located at (p, v), v<q, v can be chosen freely by the player.
3.If q=1 and p>1, one new bomb will be produced, located at (u, q), u<p, u can be chosen freely by the player.

If two bombs located at the same position or a bomb located at (1, 1), they will be exploded automatically without producing new bombs.
Two players play in turn, until one player cannot explode the bombs and loses the game.
John always plays first.
Now, we’ll give you an initial situation, and you should tell us who will win at last. Assume John and Jack are smart enough, and they always do the best choice.

Input
There are several test cases, each one begins with two integers n and m, 0<n, m<=50, represents the number of rows and columns. Following by an n*m grid, describing the initial situation, ‘#’ indicates bomb.
The input is terminated by n=m=0.

Output
For each test case, output one line, the name of the winner.

Sample Input
2 2
.#
..
2 2
.#
.#
0 0

Sample Output
John
Jack

Source
2009 Multi-University Training Contest 8 - Host by BJNU

Recommend
gaojie

这题A的我好辛苦，题目所说的1是我们输入的矩阵的0，即（1,1）这个位置是我们输入的矩阵的（0,0）位置，晕死，我直接想死了。
[cpp]
#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<memory.h>
using namespace std;
int sg[60][60];
char str[60][60];
int num[600];
int dfs(int x,int y)
{
    memset(num,-1,sizeof(num));
    for(int i=0;i<x;i++)
    for(int j=0;j<y;j++)
    {
        num[sg[i][y]^sg[x][j]]=1;
    }
    for(int i=0;;i++)
    if(num[i]==-1) return i;
}
int main()
{
    for(int i=0;i<=50;i++)
    sg[0][i]=sg[i][0]=i;
    for(int i=1;i<=50;i++)
    for(int j=1;j<=50;j++)
    sg[i][j]=dfs(i,j);
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0&&m==0) break;
        int ans=0;
        for(int i=0;i<n;i++)
        {
            scanf("%s",str[i]);
            for(int j=0;j<m;j++)
            {
                if(str[i][j]=='#')
                {
                    ans^=sg[i][j];
                }
            }
        }
        if(ans) puts("John");
        else puts("Jack");
    }
}

补充：软件开发 , C++ ,