180 lines
3.1 KiB
C
180 lines
3.1 KiB
C
#include "block.h"
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#include "draw.h"
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#include "button.h"
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char base[MAX_X][MAX_Y] = {0}; //x*y //0为空 1为下落完成 2为正在下落
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enum BLK_TP type = none;
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void blockInit()
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{
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base[0][0] = 1;
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base[0][1] = 1;
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base[0][2] = 1;
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}
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void blockDestroy()
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{
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memset(base, 0, sizeof(base));
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}
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void genPiece()
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{
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char i, j;
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//需要一点随机性
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int random = rand() % 4;
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int len = 2 + rand() % 3;
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//若存在尚未下落完成的方块则打断
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for(i = 0; i < MAX_X; i++)
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for(j = 0; j < MAX_Y; j++)
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if(base[i][j] == 2)
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break;
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//当两个方块的情况
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if(len == 2)
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{
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if(random % 2 == 0)
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{
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base[4][7] = 2; //
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base[4][8] = 2; //
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}
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else
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{
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base[3][7] = 2; ////
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base[4][7] = 2;
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}
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}
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//当三个方块的情况
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if(len == 3)
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{
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if(random == 0)
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{
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base[4][7] = 2;
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base[4][8] = 2;
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base[4][9] = 2;
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}
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else if(random == 1)
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{
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base[3][7] = 2;
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base[4][7] = 2;
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base[5][7] = 2;
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}
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else if(random == 2)
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{
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base[3][8] = 2;
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base[3][7] = 2;
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base[4][7] = 2;
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}
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else if (random == 3)
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{
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base[4][8] = 2;
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base[3][7] = 2;
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base[4][7] = 2;
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}
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}
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//当四个方块的情况
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if(len == 3)
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{
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base[3][7] = 2;
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base[4][7] = 2;
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base[3][8] = 2;
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base[4][8] = 2;
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}
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}
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//动作是否在屏幕内
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unsigned char isIegal(char i, char j)
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{
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if(i >= 0 && i < MAX_X && j >= 0 && j < MAX_Y && base[i][j] == 0)
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return 1;
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return 0;
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}
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//方块绘制
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void drawBlock()
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{
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char i, j;
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for(i = 0; i < 8; i++)
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for(j = 0; j < 8; j++)
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if(base[i][j] != 0)
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placeIMG_BLOCK(i, j);
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else
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place_VOID_BLOCK(i, j);
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}
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//方块下落与合并
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void dropPiece()
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{
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char i, j, flag = 0;
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//检测是否触底
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for(i = 0; i < 8; i++)
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{
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for(j = 0; j < 8; j++)
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{
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if(flag == 1)
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break;
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if(base[i][j] == 2)
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{
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//若触底,做标记
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if(flag == 0 && (base[i][j - 1] == 1 || j == 0))
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{
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flag = 1;
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break;
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}
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}
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}
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if(flag == 1)
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break;
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}
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//转换嵌套
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for(i = 0; i < 8; i++)
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{
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for(j = 0; j < 8; j++)
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{
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if(base[i][j] == 2)
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{
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//无标记则下落
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if(flag == 0 && isIegal(i, j - 1))
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{
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base[i][j - 1] = 2;
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base[i][j] = 0;
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}
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//有标记则转换
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else if(flag == 1)
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base[i][j] = 1;
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}
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}
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}
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}
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//方块平移
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void movePiece(enum OPR opr)
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{
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char i, j;
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for(i = 0; i < 8; i++)
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for(j = 0; j < 8; j++)
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if(base[i][j] == 2)
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{
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if(opr == left && isIegal(i - 1, j))
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{
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base[i - 1][j] = 2;
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base[i][j] = 0;
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}
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else if(opr == left && isIegal(i + 1, j))
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{
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base[i + 1][j] = 2;
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base[i][j] = 0;
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}
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}
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}
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