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四色三消游戏算法

时间:2015-01-16 16:42:27      阅读:584      评论:0      收藏:0      [点我收藏+]

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四色三消游戏算法


下面是用python写的四色三消游戏算法,很容易改成更多颜色和行列的。基本思路就是3个一样的diamonds连在一起就可以消除。废话不说,上代码:

#!/usr/bin/python
#-*- coding: UTF-8 -*-
#======================================================================
import os 
import sys
import getopt
import time
import random

#======================================================================
# color output
#
def perror(s):
    print ‘\033[31m[ERROR] %s\033[31;m‘ % (s)


def pinfo(s):
    print ‘\033[32m[INFO] %s\033[32;m‘ % (s)


def pwarn(s):
    print ‘\033[33m[WARN] %s\033[33;m‘ % (s)

#----------------------------------------------------------
def red():
    print ‘\033[31mA\033[31;m‘,

def yellow():
    print ‘\033[33mB\033[33;m‘,

def blue():
    print ‘\033[34mC\033[34;m‘,

def green():
    print ‘\033[32mD\033[32;m‘,

def one():
    print ‘\033[31m1\033[31;m‘,

def zero():
    print ‘O‘,
    

def pout(C):
    if C==‘A‘:
        red()
    elif C==‘B‘:
        yellow()
    elif C==‘C‘:
        blue()
    elif C==‘D‘:
        green()
    else:
        zero()


#==========================================================
# 4 colors and 3 crash for 5x7 diamonds table
CRASH3_COLS = 7
CRASH3_ROWS = 5
CRASH3_COLORS = [‘O‘,‘A‘,‘B‘,‘C‘,‘D‘]

CRASH3_DIAMONDS_TABLE = [
    [0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0],
]


CRASH3_DIAMONDS_RESULT = [
    [0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0],
    [0,0,0,0,0,0,0],
]


def crash3_print_colors(table):
    for row in range(0, CRASH3_ROWS, 1):
        for col in range(0, CRASH3_COLS, 1):
            color = table[row][col]
            pout(CRASH3_COLORS[color])
        print ‘‘


def crash3_print_values(table):
    for row in range(0, CRASH3_ROWS, 1):
        for col in range(0, CRASH3_COLS, 1):
            V = table[row][col]
            if V == 1:
                one()
            else:
                zero()
        print ‘‘


def crash3_reset_value(table, V):
    for row in range(0, CRASH3_ROWS, 1):
        for col in range(0, CRASH3_COLS, 1):
            table[row][col] = V


def crash3_init_table(table):
    print("\n----------------------\ncrash3_init_table:\n----------------------")
    color = random.randint(1, 4)
    for row in range(0, CRASH3_ROWS, 1):
        for col in range(0, CRASH3_COLS, 1):
            color = random.randint(1, 4)
            table[row][col] = color


def crash3_on_cell(table, result, row, col):
    if col < CRASH3_COLS - 2:
        (a,b,c) = (table[row][col], table[row][col+1], table[row][col+2])
        if a==b and b==c:
            result[row][col] = 0;
            result[row][col+1] = 0;
            result[row][col+2] = 0;

    if row < CRASH3_ROWS - 2:
        (a,b,c) = (table[row][col], table[row+1][col], table[row+2][col])
        if a==b and b==c:
            result[row][col] = 0;
            result[row+1][col] = 0;
            result[row+2][col] = 0;


def crash3_on_trigger(table, result):
    print("\n----------------------\ncrash3_on_trigger:\n----------------------")
    for row in range(0, CRASH3_ROWS, 1):
        for col in range(0, CRASH3_COLS, 1):
            crash3_on_cell(table, result, row, col)


#==========================================================
# main() entry
if __name__ == "__main__":

    pinfo("crash linked 3 diamonds.\ncopyright by cheungmine, all rights reserved!")

    crash3_init_table(CRASH3_DIAMONDS_TABLE)

    crash3_print_colors(CRASH3_DIAMONDS_TABLE)

    crash3_reset_value(CRASH3_DIAMONDS_RESULT, 1)

    crash3_on_trigger(CRASH3_DIAMONDS_TABLE, CRASH3_DIAMONDS_RESULT)

    crash3_print_values(CRASH3_DIAMONDS_RESULT)
    

运行结果截图,O表示消除,1表示未消除:

技术分享

实际使用中,很容易改为其他语言的。

四色三消游戏算法

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原文地址:http://blog.csdn.net/ubuntu64fan/article/details/42778691

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