SimpleCG动画示例--汉诺塔动画演示

前言

        SimpleCG的使用方法在前面已经介绍了许多，有兴趣的同学如果有去动手，制作一些简单动画应该没多大问题的。所以这次我们来演示一下简单动画。我们刚学习C语言的递归函数时，有一个经典例子相信很多同学都写过，那就是汉诺塔。那么我们今天就来写一个汉诺塔的直观动画演示。

运行程序下载bin/AnimateHannuo.zip · master · b2b160 / SimpleCG_Demo · GitCode

一、全部源码

#include "../import/include/CGBoard.h"
#include "math.h"
#ifdef _DEBUG
#pragma comment(lib,"../import/lib/SimpleCG_MDd.lib")
#else
#pragma comment(lib,"../import/lib/SimpleCG_MT.lib")
#endif#define C_FLOOR_CNT		7
#define C_FLOOR_BOTTOM  380
#define C_HAN_HEIGHT	10
#define C_HAN_SPEED		30int g_nWidth = 640;		//画面宽度
int g_nHeight= 400;		//画面高度
enum ENUM_DIRECTION
{enumDIR_NULL,enumDIR_UP,enumDIR_DOWN
};
struct tagHannuo
{int nNumber;COLORREF nColor;int nWidth;int nPosHan;int nDir;POINT ptPos;
};tagHannuo g_pHannuo[C_FLOOR_CNT];
int g_nMoving = -1;
void DrawHan()
{int i;int j=0;setlinewidth(2);for(i=0;i<3; i++ ){_line( 100 + 200 * i, 50, 100 + 200 * i,C_FLOOR_BOTTOM );_line( 20 + 200 * i, C_FLOOR_BOTTOM, 180 + 200 * i,C_FLOOR_BOTTOM );}int nIndex = 0;for(j=0;j<3;++j){nIndex = 0;for(i=C_FLOOR_CNT-1;i>=0; i-- ){if(g_pHannuo[i].nPosHan == j && i != g_nMoving){setfillcolor(g_pHannuo[i].nColor);_solidrectangle( 100 +200 * g_pHannuo[i].nPosHan - g_pHannuo[i].nWidth/2, C_FLOOR_BOTTOM - nIndex * C_HAN_HEIGHT - C_HAN_HEIGHT, 100 +200 * g_pHannuo[i].nPosHan + g_pHannuo[i].nWidth/2, C_FLOOR_BOTTOM - nIndex * C_HAN_HEIGHT);++nIndex;}}}
}
void DrawMoving()
{if(g_nMoving>=0){setfillcolor(g_pHannuo[g_nMoving].nColor);_solidrectangle( g_pHannuo[g_nMoving].ptPos.x, g_pHannuo[g_nMoving].ptPos.y, g_pHannuo[g_nMoving].ptPos.x + g_pHannuo[g_nMoving].nWidth, g_pHannuo[g_nMoving].ptPos.y+ C_HAN_HEIGHT);}
}
void DrawAll()
{ClearDevice();DrawHan();DrawMoving();ReflushWindow();
}
void Moving( int nItem, int nFrom, int nTo )
{g_nMoving = nItem;g_pHannuo[nItem].ptPos.x = 100 +200 * nFrom - g_pHannuo[nItem].nWidth/2;for( g_pHannuo[nItem].ptPos.y = C_FLOOR_BOTTOM - C_FLOOR_CNT * C_HAN_HEIGHT; IsShowingWindow()&&g_pHannuo[nItem].ptPos.y>40; g_pHannuo[nItem].ptPos.y-=10 ){DrawAll();Sleep(C_HAN_SPEED);}int nXStep = (nTo - nFrom) * 5;int nDest = 100 +200 * nTo - g_pHannuo[nItem].nWidth/2;for( g_pHannuo[nItem].ptPos.x = 100 +200 * nFrom - g_pHannuo[nItem].nWidth/2; IsShowingWindow()&&abs(g_pHannuo[nItem].ptPos.x-nDest)>5; g_pHannuo[nItem].ptPos.x+=nXStep ){DrawAll();Sleep(C_HAN_SPEED);}g_pHannuo[nItem].ptPos.x = 100 +200 * nTo - g_pHannuo[nItem].nWidth/2;for( g_pHannuo[nItem].ptPos.y = 40; IsShowingWindow()&&g_pHannuo[nItem].ptPos.y<C_FLOOR_BOTTOM - C_FLOOR_CNT * C_HAN_HEIGHT; g_pHannuo[nItem].ptPos.y+=10 ){DrawAll();Sleep(C_HAN_SPEED);}g_nMoving = -1;g_pHannuo[nItem].nPosHan = nTo;
}
void MoveHan( int nFloor, int nFrom, int nTo, int nMiddle )
{if( nFloor == 1 ){Moving( nFloor-1, nFrom-1, nTo-1);return;}MoveHan( nFloor-1, nFrom, nMiddle, nTo );Moving( nFloor-1, nFrom-1, nTo-1);MoveHan( nFloor-1, nMiddle, nTo, nFrom );
}
void DrawProcess()
{bool bIsRunning = true;int i;srand(GetTickCount());for(i=0;i<C_FLOOR_CNT; i++ ){g_pHannuo[i].nNumber=i+1;g_pHannuo[i].nColor = RGB(rand()%200,rand()%200,rand()%200);g_pHannuo[i].nDir = enumDIR_NULL;g_pHannuo[i].nPosHan = 0;g_pHannuo[i].nWidth = 20*(i+1);}MoveHan(C_FLOOR_CNT,1,2,3);DrawAll();
}
int _tmain(int argc, _TCHAR* argv[])
{//初始化if( !ShowingBoard(g_nWidth,g_nHeight, DrawProcess))return 1;//关闭图库CloseBoard();return 0;
}

对于写过汉诺塔的同学来说，程序逻辑应该没什么难度，就是在递归程序上增加了动画过程。

二、演示效果

对于5层来说是不难的，但递归对于层数增加所带来的时间消耗是呈指数增加的，所以通过动画来观察层数增加带来的时间消耗非常直观。在原始的汉诺塔里是64层，要移完所有的层数将会世界末日，因为即便到世界的尽头也无法完成。有兴趣的同学可以把代码输入并把层数加大看看。

三、代码下载

汉诺塔演示源代码

AnimateHannuo · master · b2b160 / SimpleCG_Demo · GitCode

库安装方法如下

SimpleCG库安装使用_b2b160的博客-CSDN博客