C++基于EasyX实现简单扫雷游戏

本文实例为大家分享了C++ EasyX实现简单扫雷游戏的具体代码，供大家参考，具体内容如下

【实现代码】

1. #include <cmath>
2. #include <time.h>
3. #include <easyx.h>
4. #include <conio.h>
5. using namespace std;
6. #define Size 500 //定义窗口大小
7. #define SquareSize 50 //定义格子大小
8. #define BackGroundColor LIGHTGRAY //定义背景颜色
9. const int N = Size / SquareSize; //设置地图大小为N * N
10. const int Num = N * N / 5; //设置总地雷数
11. int num = Num, all = N * N - Num; //剩余地雷数和未格子访问数
12. int Map[N + 10][N + 10]; //地图表示(i, j)旁边的地雷数目，等于-1表示该位置为地雷
13. bool vis[N + 10][N + 10]; //标记是否(i, j)被访问
14. bool vis_lm[N + 10][N + 10]; //标记(i, j)是否被标记为地雷
15. int dx[] = { 0, 0, 1, -1, 1, -1, 1, -1 };
16. int dy[] = { 1, -1, 0, 0, 1, -1, -1, 1 };
17. void Init() {
18. initgraph(Size, Size);
19. setbkcolor(BackGroundColor);
20. settextcolor(YELLOW);
21. settextstyle(25, 0, _T("宋体"));
22. cleardevice();
23. }
24. void BuildGameMap() { //构建地图
25. srand((unsigned int)time(NULL)); //产生随机数
26. for (int i = 1; i <= Num; ++i) {
27. int x = rand() % N + 1, y = rand() % N + 1;
28. while (Map[x][y] == -1) {
29. x = rand() % N + 1, y = rand() % N + 1;
30. }
31. Map[x][y] = -1;
32. }
33. //计算每个格子旁边的地雷数目
34. for (int i = 1; i <= N; ++i) {
35. for (int j = 1; j <= N; ++j) {
36. if (Map[i][j] == -1) continue;
37. for (int k = 0; k < 8; ++k) {
38. if (Map[i + dx[k]][j + dy[k]] == -1) ++Map[i][j];
39. }
40. }
41. }
42. }
43. void InitGameGraph() { //地图初始化
44. memset(Map, 0, sizeof(Map));
45. memset(vis, false, sizeof(vis));
46. memset(vis_lm, false, sizeof(vis_lm));
47. num = Num, all = N * N - Num;
48. BuildGameMap();
49. cleardevice(); //清屏
50. setfillcolor(WHITE);
51. for (int i = 0; i <= Size; i += SquareSize) {
52. line(i, 0, i, Size);
53. }
54. for (int i = 0; i <= Size; i += SquareSize) {
55. line(0, i, Size, i);
56. }
57. }
58. int Translate_X(int x) { //左边变换
59. return x * SquareSize + SquareSize / 2;
60. }
61. void GameOver(bool tag) {
62. RECT r = { 0, 0, Size - 1, Size - 1 };
63. if (tag) {
64. drawtext(_T("Win"), &r, DT_CENTER | DT_VCENTER | DT_SINGLELINE);
65. }
66. else {
67. //游戏失败画出所有地雷
68. setfillcolor(RED);
69. for (int i = 1; i <= N; ++i) {
70. for (int j = 1; j <= N; ++j) {
71. if (Map[i][j] == -1) {
72. if (vis_lm[i][j]) continue; //该地雷已被玩家标记，跳过
73. outtextxy(Translate_X(i - 1), Translate_X(j - 1), 'X'); //否则显示出来
74. }
75. else if (vis_lm[i][j]) { //玩家标记了错误的地雷，标红
76. fillcircle(Translate_X(i - 1), Translate_X(j - 1), SquareSize / 2 - 10);
77. }
78. }
79. }
80. drawtext(_T("Failed"), &r, DT_CENTER | DT_VCENTER | DT_SINGLELINE);
81. }
82. }
83. void Help() { //调试函数（其实是挂）
84. for (int i = 1; i <= N; ++i) {
85. for (int j = 1; j <= N; ++j) {
86. if (Map[i][j] == -1) {
87. outtextxy(Translate_X(i - 1), Translate_X(j - 1), 'X');
88. }
89. }
90. }
91. }
92. void StartGame() {
93. Init();
94. while (true) {
95. InitGameGraph();
96. //Help();
97. bool IF_WIN = true;
98. while (all || num) { //胜利条件是所有非地雷格子均被访问且所有地雷均被标记
99. MOUSEMSG msg = GetMouseMsg(); //获取用户鼠标信息
100. int x = ceil(msg.x * 1.0 / SquareSize * 1.0), y = ceil(msg.y * 1.0 / SquareSize * 1.0);
101. if (vis[x][y]) continue; //该格已被访问，跳过
102. if (msg.mkLButton) { //鼠标左键按下
103. if (Map[x][y] == -1) { //如果是地雷那么游戏结束
104. IF_WIN = false;
105. break;
106. }
107. outtextxy(Translate_X(x - 1), Translate_X(y - 1), Map[x][y] + '0');
108. --all;
109. vis[x][y] = true;
110. }
111. else if (msg.mkRButton) { //鼠标右键按下
112. if (!vis_lm[x][y]) { //如果该格子已被标记，消除标记
113. vis_lm[x][y] = true;
114. if (Map[x][y] == -1) --num;
115. solidcircle(Translate_X(x - 1), Translate_X(y - 1), SquareSize / 2 - 10);
116. }
117. else if (vis_lm[x][y]) { //否则标记
118. vis_lm[x][y] = false;
119. if (Map[x][y] == -1) ++num;
120. clearrectangle((x - 1) * SquareSize + 1, (y - 1) * SquareSize + 1, x * SquareSize - 1, y * SquareSize - 1);
121. }
122. }
123. }
124. GameOver(IF_WIN);
125. int t = _getch();
126. }
127. }
128. int main() {
129. StartGame();
130. return 0;
131. }

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持我们。

原文链接：https://blog.csdn.net/weixin_44211980/article/details/104543819