C语言【微项目10】—棋盘覆盖问题[递归到基本四格单位时填充]（采用分治法实现）
【TDTX】
【MyCoverBoardA.c】把棋盘的填充视为：【排除缺陷式】位置填充；
【MyCoverBoardB.c】把棋盘的填充视为：【图形卡位式】位置填充；
【总结】两者最终的结果完全相同，但视觉效果不同。
一、MyCoverBoardA.c
#include <stdio.h>
#include <stdlib.h>
int qa,qb;
int mpow(int n,int k)
{
int i;
int t = n;
for(i = 1; i < k;i++)
{
t = t * n;
}
return t;
}
void generateBoard(int tr,int tc,int dr,int dc,int size,int** b,int* t1)
{
int hs = size / 2;
if(size == 2)
{
if( dr == tr && dc == tc )
{
//残缺位置在4格左上角
b[tr][tc+1] = *t1;
b[tr+1][tc] = *t1;
b[tr+1][tc+1] = *t1;
(*t1)++;
}
if( dr== tr && dc == tc + 1)
{
//残缺位置在4格右上角
b[tr][tc] = *t1;
b[tr+1][tc] = *t1;
b[tr+1][tc+1] = *t1;
(*t1)++;
}
if( dr == tr + 1 && dc == tc)
{
//残缺位置在4格左下角
b[tr][tc] = *t1;
b[tr][tc+1] = *t1;
b[tr+1][tc+1] = *t1;
(*t1)++;
}
if( dr == tr + 1 && dc == tc +1)
{
//残缺位置在4格右下角
b[tr][tc] = *t1;
b[tr][tc+1] = *t1;
b[tr+1][tc] = *t1;
(*t1)++;
}
/*
for(int i = 0;i < 2*size;i++)
{
for(int j = 0;j < 2*size;j++)
{
printf("%d\t",b[i][j]);
}
puts("");
}
*/
//printf("\ntr:%d,tc:%d----------------------\n",tr,tc);
b[qa][qb] = 0;
return;
}
else
{
if(dr < tr + hs && dc < tc + hs)
{
int tdr,tdc;
//残缺在第二象限
b[tr+hs-1][tc+hs] = *t1;
b[tr+hs][tc+hs-1] = *t1;
b[tr+hs][tc+hs] = *t1;
(*t1)++;
generateBoard(tr,tc,dr,dc,hs,b,t1);
tdr = tr + hs - 1;
tdc = tc + hs;
generateBoard(tr,tc+hs,tdr,tdc,hs,b,t1);
tdr = tr + hs;
tdc = tc + hs - 1;
generateBoard(tr+hs,tc,tdr,tdc,hs,b,t1);
tdr = tr + hs;
tdc = tc + hs;
generateBoard(tr+hs,tc+hs,tdr,tdc,hs,b,t1);
}
if(dr < tr + hs && dc >= tc + hs)
{
int tdr,tdc;
//残缺在第一象限
b[tr+hs-1][tc+hs-1] = *t1;
b[tr+hs][tc+hs-1] = *t1;
b[tr+hs][tc+hs] = *t1;
(*t1)++;
generateBoard(tr,tc+hs,dr,dc,hs,b,t1);
tdr = tr + hs - 1;
tdc = tc + hs - 1;
generateBoard(tr,tc,tdr,tdc,hs,b,t1);
tdr = tr + hs;
tdc = tc + hs - 1;
generateBoard(tr+hs,tc,tdr,tdc,hs,b,t1);
tdr = tr + hs;
tdc = tc + hs;
generateBoard(tr+hs,tc+hs,tdr,tdc,hs,b,t1);
}
if(dr >= tr + hs && dc < tc + hs)
{
int tdr,tdc;
//残缺在第三象限
b[tr+hs-1][tc+hs-1] = *t1;
b[tr+hs-1][tc+hs] = *t1;
b[tr+hs][tc+hs] = *t1;
(*t1)++;
generateBoard(tr+hs,tc,dr,dc,hs,b,t1);
tdr = tr + hs - 1;
tdc = tc + hs - 1;
generateBoard(tr,tc,tdr,tdc,hs,b,t1);
tdr = tr + hs - 1;
tdc = tc + hs;
generateBoard(tr,tc+hs,tdr,tdc,hs,b,t1);
tdr = tr + hs;
tdc = tc + hs;
generateBoard(tr+hs,tc+hs,tdr,tdc,hs,b,t1);
}
if(dr >= tr + hs && dc >= tc + hs)
{
int tdr,tdc;
//残缺在第四象限
b[tr+hs-1][tc+hs-1] = *t1;
b[tr+hs-1][tc+hs] = *t1;
b[tr+hs][tc+hs-1] = *t1;
(*t1)++;
generateBoard(tr+hs,tc+hs,dr,dc,hs,b,t1);
tdr = tr+hs-1;
tdc = tc+hs-1;
generateBoard(tr,tc,tdr,tdc,hs,b,t1);
tdr = tr+hs-1;
tdc = tc+hs;
generateBoard(tr,tc+hs,tdr,tdc,hs,b,t1);
tdr = tr+hs;
tdc = tc+hs-1;
generateBoard(tr+hs,tc,tdr,tdc,hs,b,t1);
}
}
}
int main()
{
//左下残缺 右下残缺 左上残缺 右上残缺
//* * * * * *
// * * * * * *
int k;
printf("输入棋盘2^k×2^k的k值：");
scanf("%d",&k);
int size = mpow(2,k);
//printf("size = %d\n",size);
int *b[size]; //生成棋盘矩阵
for(int i = 0;i < size;i++)
{
b[i] = (int*)malloc(sizeof(int)*size);
}
for(int i = 0;i < size;i++)
{
for(int j = 0;j < size;j++)
{
b[i][j] = -1;
}
//puts("");
}
printf("输入残缺位置（行 列）：");
scanf("%d %d",&qa,&qb);
int dr,dc;
dr = qa;
dc = qb;
int t1 = 1;
if((qa < size && qa >= 0) && (qb < size && qb >= 0))
{
b[qa][qb] = 0;
generateBoard(0,0,dr,dc,size,b,&t1);
for(int i = 0;i < size;i++)
{
for(int j = 0;j < size;j++)
{
printf("%2d ",b[i][j]);
}
puts("");
}
}
else
{
printf("残缺位置不在棋盘上，位置不合法！");
}
for(int i = 0;i < size;i++)
{
free(b[i]);
}
system("pause");
return 0;
}
二、 MyCoverBoardB.c
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
int qa,qb;
int mpow(int n,int k)
{
int i;
int t = n;
for(i = 1; i < k;i++)
{
t = t * n;
}
return t;
}
void generateBoard(int tr,int tc,int dr,int dc,int size,int** b)
{
int hs = size / 2;
if(size == 2)
{
if( dr == tr && dc == tc )
{
//残缺位置在4格左上角
//用3号覆盖
b[tr][tc+1] = 3;
b[tr+1][tc] = 3;
b[tr+1][tc+1] = 3;
}
if( dr== tr && dc == tc + 1)
{
//残缺位置在4格右上角
//用4号覆盖
b[tr][tc] = 4;
b[tr+1][tc] = 4;
b[tr+1][tc+1] = 4;
}
if( dr == tr + 1 && dc == tc)
{
//残缺位置在4格左下角
//用1号覆盖
b[tr][tc] = 1;
b[tr][tc+1] = 1;
b[tr+1][tc+1] = 1;
}
if( dr == tr + 1 && dc == tc +1)
{
//残缺位置在4格右下角
//用2号覆盖
b[tr][tc] = 2;
b[tr][tc+1] = 2;
b[tr+1][tc] = 2;
}
/*
for(int i = 0;i < 2*size;i++)
{
for(int j = 0;j < 2*size;j++)
{
printf("%d\t",b[i][j]);
}
puts("");
}
*/
//printf("\ntr:%d,tc:%d----------------------\n",tr,tc);
b[qa][qb] = 0;
return;
}
else
{
if(dr < tr + hs && dc < tc + hs)
{
int tdr,tdc;
//残缺在第二象限
b[tr+hs-1][tc+hs] = 3;
b[tr+hs][tc+hs-1] = 3;
b[tr+hs][tc+hs] = 3;
generateBoard(tr,tc,dr,dc,hs,b);
tdr = tr + hs - 1;
tdc = tc + hs;
generateBoard(tr,tc+hs,tdr,tdc,hs,b);
tdr = tr + hs;
tdc = tc + hs - 1;
generateBoard(tr+hs,tc,tdr,tdc,hs,b);
tdr = tr + hs;
tdc = tc + hs;
generateBoard(tr+hs,tc+hs,tdr,tdc,hs,b);
}
if(dr < tr + hs && dc >= tc + hs)
{
int tdr,tdc;
//残缺在第一象限
b[tr+hs-1][tc+hs-1] = 4;
b[tr+hs][tc+hs-1] = 4;
b[tr+hs][tc+hs] = 4;
generateBoard(tr,tc+hs,dr,dc,hs,b);
tdr = tr + hs - 1;
tdc = tc + hs - 1;
generateBoard(tr,tc,tdr,tdc,hs,b);
//generateBoard(tr,tc+hs,dr,dc,hs,b);
tdr = tr + hs;
tdc = tc + hs - 1;
generateBoard(tr+hs,tc,tdr,tdc,hs,b);
tdr = tr + hs;
tdc = tc + hs;
generateBoard(tr+hs,tc+hs,tdr,tdc,hs,b);
}
if(dr >= tr + hs && dc < tc + hs)
{
int tdr,tdc;
//残缺在第三象限
b[tr+hs-1][tc+hs-1] = 1;
b[tr+hs-1][tc+hs] = 1;
b[tr+hs][tc+hs] = 1;
generateBoard(tr+hs,tc,dr,dc,hs,b);
tdr = tr + hs - 1;
tdc = tc + hs - 1;
generateBoard(tr,tc,tdr,tdc,hs,b);
tdr = tr + hs - 1;
tdc = tc + hs;
generateBoard(tr,tc+hs,tdr,tdc,hs,b);
//generateBoard(tr+hs,tc,dr,dc,hs,b);
tdr = tr + hs;
tdc = tc + hs;
generateBoard(tr+hs,tc+hs,tdr,tdc,hs,b);
}
if(dr >= tr + hs && dc >= tc + hs)
{
int tdr,tdc;
//残缺在第四象限
b[tr+hs-1][tc+hs-1] = 2;
b[tr+hs-1][tc+hs] = 2;
b[tr+hs][tc+hs-1] = 2;
generateBoard(tr+hs,tc+hs,dr,dc,hs,b);
tdr = tr+hs-1;
tdc = tc+hs-1;
generateBoard(tr,tc,tdr,tdc,hs,b);
tdr = tr+hs-1;
tdc = tc+hs;
generateBoard(tr,tc+hs,tdr,tdc,hs,b);
tdr = tr+hs;
tdc = tc+hs-1;
generateBoard(tr+hs,tc,tdr,tdc,hs,b);
//generateBoard(tr+hs,tc+hs,dr,dc,hs,b);
}
}
}
int main()
{
//左下残缺 右下残缺 左上残缺 右上残缺
//1 1 2 2 3 4
// 1 2 3 3 4 4
//把基础的四格情况分为四种样式的图形，去图形式的填充
int k;
printf("输入棋盘2^k×2^k的k值：");
scanf("%d",&k);
int size = mpow(2,k);
//printf("size = %d\n",size);
int *b[size]; //生成棋盘矩阵
for(int i = 0;i < size;i++)
{
b[i] = (int*)malloc(sizeof(int)*size);
}
for(int i = 0;i < size;i++)
{
for(int j = 0;j < size;j++)
{
b[i][j] = -1;
}
//puts("");
}
printf("输入残缺位置（行 列）：");
scanf("%d %d",&qa,&qb);
int dr,dc;
dr = qa;
dc = qb;
if((qa < size && qa >= 0) && (qb < size && qb >= 0))
{
b[qa][qb] = 0;
generateBoard(0,0,dr,dc,size,b);
for(int i = 0;i < size;i++)
{
for(int j = 0;j < size;j++)
{
printf("%d ",b[i][j]);
}
puts("");
}
}
else
{
printf("残缺位置不在棋盘上，位置不合法！");
}
for(int i = 0;i < size;i++)
{
free(b[i]);
}
system("pause");
return 0;
}
MyCoverBoardA.c-运行结果示例
1.k为1
1.1缺陷在1 0位置
1.2缺陷在0 0位置
2.k为2
2.1缺陷在1 2位置
2.2缺陷在2 3位置
3.k为3
3.1缺陷在5 6位置
3.1缺陷在4 2位置
4.k为4
4.1 3.1缺陷在12 9位置
MyCoverBoardB.c-运行结果示例
1.k为1
1.2缺陷在0 0位置
1.1缺陷在0 1位置
1.1缺陷在1 0位置
1.2缺陷在1 1位置
2.k为2
2.1缺陷在1 2位置
2.2缺陷在2 3位置
3.k为3
3.1缺陷在5 6位置
3.1缺陷在4 2位置
4.k为4
4.1 3.1缺陷在12 9位置
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-----------------------------------------------------【C语言—微项目—自编练习】----------------------------------------------------------
----------------------------------------------------------------【TDTX】--------------------------------------------------------------------------
