种菜建模
[編輯] [转简体] （简体译文）
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作者：zouhaiyuan
| 分類：【算法】種菜算法
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概要
种菜建模 有两种方案
1）横排
2)竖排
注意:
横排与竖排可以通过地块及障碍物坐标变换来实现
每种方案都采用如下4个步骤
step1.求取宽度方向能放几排，每排如何组合
step2.求取长度方向组合，即一排中能放什么规格的箱子
step3.避开障碍物
step4.求取种植面积
正文
#include <stdio.h>
typedef struct _tagBlock
{
int left;
int bottom;
int length;
int width;
}BLOCK;
typedef struct _tagFarmland
{
int length;
int width;
BLOCK stBlock;
}FARMLAND;
//箱子宽度方向组合利用率
typedef struct _tagBoxWidthAvailability
{
int width1;
int width2;
double availability;
}BOX_WIDTH_AVAILABILITY;
typedef struct _tagRowClass
{
BOX_WIDTH_AVAILABILITY stWidthAvailability;
int nRowNum;
}ROW_CLASS;
typedef struct _tagColClass
{
int nLenSpec;
int nColNum;
}COL_CLASS;
typedef struct _tagRowRect
{
int left;
int top;
int right;
int bottom;
int halfTop;//半行处的top
int width1;//上半行箱子宽度
int width2;//下半行箱子宽度
int realBoxNum1;//上半行箱子数量
int realBoxNum2;//下半行箱子数量
int blocked1;//上半行是否受阻，0--未障碍,1--受障碍
int blocked2;//下半行是否受阻，0--未障碍,1--受障碍
int ailseBlocked;//所属走道是否阻塞,0--未受阻，1--受阻，注意，走道在每行的下侧
int boxClassIndex;//改行采用的箱子组合类型序号
}ROW_RECT;
int GetArea(int p_pnLenBoxSpecs[], int p_nNumLenSpecs, int p_pnWidthBoxSpecs[], int p_nNumWidthSpecs, int p_nAisle, FARMLAND p_stFarmland)
{
//将长度规格和宽度规格从大到小排好序
//注意,每种长度规格都有其对应的全部宽度规格，即总规格数量为 长度规格数量*宽度规格数量
//int pnLenBoxSpecs[] = { 80,120,200,160};
//int pnWidthBoxSpecs[] = {80,40};
int* pnLenBoxSpecs = new int[p_nNumLenSpecs];
int* pnWidthBoxSpecs = new int[p_nNumWidthSpecs];
int nNumLenBoxSpecs = p_nNumLenSpecs;//sizeof(pnLenBoxSpecs)/ sizeof(int);
int nNumWidthBoxSpecs = p_nNumWidthSpecs;//sizeof(pnWidthBoxSpecs)/ sizeof(int);
for (int i = 0; i < nNumLenBoxSpecs; i++)
{
pnLenBoxSpecs[i] = p_pnLenBoxSpecs[i];
}
for (int i = 0; i < nNumWidthBoxSpecs; i++)
{
pnWidthBoxSpecs[i] = p_pnWidthBoxSpecs[i];
}
printf("种植箱规格:\r\n");
for (int i = 0; i < nNumWidthBoxSpecs; i++)
{
for (int j = 0; j < nNumLenBoxSpecs; j++)
{
printf("%d x %d\r\n", pnLenBoxSpecs[j], pnWidthBoxSpecs[i]);
}
}
printf("\r\n");
FARMLAND stFarmland = p_stFarmland;//{ 1080,550,{100,200,170,140} };
printf("地块 长 %d, 宽 %d\r\n", stFarmland.length, stFarmland.width);
printf("管线井 左侧 %d,底部 %d, 长 %d, 宽 %d\r\n", stFarmland.stBlock.left, stFarmland.stBlock.bottom, stFarmland.stBlock.length, stFarmland.stBlock.width);
int aisle = p_nAisle;//30;
printf("过道宽度 %d\r\n\r\n", aisle);
//方案一，横排
//根据种植要求，种植箱长边至少一侧预留通道,
// 那么，最大利用率就是仅仅一侧留通道,即箱子A的长边和箱子B的长边紧紧挨着。
// 此时宽度方向的利用率为 (箱子A的宽度+箱子B的宽度)/(箱子A的宽度+箱子B的宽度+过道宽度)
// 由此可知，要想宽度方向利用率高，那箱子的宽度就要尽量的大
//
// 种植箱宽度方向未对过道作要求，因此箱子的宽度侧紧紧挨着，此时利用率最大
//
// 进园后需要横向和纵向过道
//
// 种植示意图如下:
//
//***********************************************************************************
//* 过道 *
//* ________________________________________________________________________ *
//*过道 | | | | |可*
//* |____________________|___________________|___________________|___________|能*
//入 | | | | |空*
//口 |____________________|___________________|___________________|___________|隙*
//* *
//* 过道 *
//* ________________________________________________________________________ *
//*过道 | | | | | *
//* |____________________|___________________|___________________|___________| *
//* | | | | | *
//* |____________________|___________________|___________________|___________| *
//* *
//* 过道 *
//* ________________________________________________________________________ *
//*过道 | | | | | *
//* |____________________|___________________|___________________|___________| *
//* | | | | | *
//* |____________________|___________________|___________________|___________| *
//* *
//* 过道 *
//* ________________________________________________________________________ *
//*过道 | | | | | *
//* |____________________|___________________|___________________|___________| *
//* | | | | | *
//* |____________________|___________________|___________________|___________| *
//* *
//* 过道 *
//* ________________________________________________________________________ *
//*过道 | | | | | *
//* |____________________|___________________|___________________|___________| *
//* |____________________|___________________|___________________|___________| *
// * *
//* 过道 *
//* *
//* 可能空隙 *
// ***********************************************************************************
//
// 如上图，将两个箱子长度方向紧紧靠在一起，又将四个箱子宽度方向靠在一起，形成一行
// 那么如果有 N 行的话，那么与之平行的过道就有N+1条过道
// 注意，最后一行可能不是两个最大宽度的箱子靠在一起，甚至可能是单个箱子成一行
// 由此，我们先将两个箱子长度靠在一起，甚至单个箱子成一行的利用率按照从大到小形成一个数组
//
printf("\r\nStep1.\r\n计算宽度方向箱子布局\r\n\r\n");
//step1.
//计算两个箱子长度靠在一起，甚至单个箱子成一行的利用率
BOX_WIDTH_AVAILABILITY* pstAvailability = new BOX_WIDTH_AVAILABILITY[nNumWidthBoxSpecs * nNumWidthBoxSpecs + nNumWidthBoxSpecs];
int nNumAvailability = 0;
int nNumWidthBoxSpecsTemp = nNumWidthBoxSpecs;
//双排箱子
while (nNumWidthBoxSpecsTemp > 0)
{
for (int i = nNumWidthBoxSpecs - nNumWidthBoxSpecsTemp; i < nNumWidthBoxSpecs; i++)
{
pstAvailability[nNumAvailability].width1 = pnWidthBoxSpecs[nNumWidthBoxSpecs - nNumWidthBoxSpecsTemp];
pstAvailability[nNumAvailability].width2 = pnWidthBoxSpecs[i];
pstAvailability[nNumAvailability].availability = (pstAvailability[nNumAvailability].width1 + pstAvailability[nNumAvailability].width2) * 1.0 / (pstAvailability[nNumAvailability].width1 + pstAvailability[nNumAvailability].width2 + aisle);
nNumAvailability++;
}
nNumWidthBoxSpecsTemp--;
}
//单排箱子
for (int i = 0; i < nNumWidthBoxSpecs; i++)
{
pstAvailability[nNumAvailability].width1 = pnWidthBoxSpecs[i];
pstAvailability[nNumAvailability].width2 = 0;
pstAvailability[nNumAvailability].availability = (pstAvailability[nNumAvailability].width1) * 1.0 / (pstAvailability[nNumAvailability].width1 + aisle);
nNumAvailability++;
}
printf("种植箱宽度组合\r\n");
for (int i = 0; i < nNumAvailability; i++)
{
printf("%d: %d - %d - %lf\r\n", i, pstAvailability[i].width1, pstAvailability[i].width2, pstAvailability[i].availability);
}
//利用率排序
int nLoop = nNumAvailability;
for (int i = 0; i < nLoop;)
{
BOX_WIDTH_AVAILABILITY stAvailability = pstAvailability[i];
for (int j = 1; j < nLoop; j++)
{
if (stAvailability.availability < pstAvailability[j].availability)
{
pstAvailability[j - 1] = pstAvailability[j];
pstAvailability[j] = stAvailability;
}
else
{
stAvailability = pstAvailability[j];
}
}
nLoop--;
}
printf("种植箱宽度组合(排序)\r\n");
for (int i = 0; i < nNumAvailability; i++)
{
printf("%d: %d - %d - %lf\r\n", i, pstAvailability[i].width1, pstAvailability[i].width2, pstAvailability[i].availability);
}
//按照利用率高的优先使用的原则布满地块宽度，求取布了多少行,及使用何种组合布置
//注意，该原则下，并不一定结果是最优解。但很接近最优解
//
/*
//下面尝试多次使用该原则找最优解,增加了代码复杂的，抛弃。。。
nLoop = nNumAvailability;//如果nLoop =1,就是优先使用最大利用率组合。不进行多次尝试。
ROW_CLASS *pstRowClass = new ROW_CLASS[nLoop*nNumAvailability];
int* pnRowClassNum = new int[nLoop];
int* pnWidthRemainder = new int[nLoop];
int* pnWidthUsed = new int[nLoop];
for (int k = 0; k < nLoop; k++)
{
int nValidWidth = stFarmland.width - aisle;//有效宽度，即去掉了第一行过道之后的剩余宽度
int nRowClassNum = 0;
int nWidthUsed = 0;
for (int i = k; i < nNumAvailability; i++)
{
int nNum = nValidWidth / (pstAvailability[i].width1 + pstAvailability[i].width2 + aisle);
if (nNum > 0)
{
pstRowClass[k* nNumAvailability + nRowClassNum].stWidthAvailability = pstAvailability[i];
pstRowClass[k * nNumAvailability + nRowClassNum].nRowNum = nNum;
nRowClassNum++;
nWidthUsed += nNum*(pstAvailability[i].width1 + pstAvailability[i].width2);
}
//剩余地块宽度
nValidWidth = nValidWidth % (pstAvailability[i].width1 + pstAvailability[i].width2 + aisle);
}
pnRowClassNum[k] = nRowClassNum;
pnWidthRemainder[k] = nValidWidth;
pnWidthUsed[k] = nWidthUsed;
printf("--组合类型数 %d ,使用的宽度%d 最后空隙 %d--\r\n", nRowClassNum, nWidthUsed,nValidWidth);
}
printf("\r\n地块宽度方向的布置\r\n");
for (int k=0;k<nLoop;k++)
{
printf("[-- %d --]\r\n", k);
for (int i = 0; i < pnRowClassNum[k]; i++)
{
printf("%d: 行数:%d 箱1宽度:%d 箱2宽度%d\r\n", i, pstRowClass[k* nNumAvailability +i].nRowNum, pstRowClass[k * nNumAvailability + i].stWidthAvailability.width1, pstRowClass[k * nNumAvailability + i].stWidthAvailability.width2);
}
printf("使用宽度 %d, 最后空隙 %d\r\n", pnWidthUsed[k], pnWidthRemainder[k]);
}
*/
//
//下面直接优先使用利用率高的箱子，不多次尝试
nLoop = 1;//如果nLoop =1,就是优先使用最大利用率组合。不进行多次尝试。
ROW_CLASS* pstRowClass = new ROW_CLASS[nLoop * nNumAvailability];
int* pnRowClassNum = new int[nLoop];
int* pnWidthRemainder = new int[nLoop];
int* pnWidthUsed = new int[nLoop];
int nValidWidth = stFarmland.width - aisle;//有效宽度，即去掉了第一行过道之后的剩余宽度
int nRowClassNum = 0;
int nWidthUsed = 0;
int nWidthRemainder = 0;
for (int i = 0; i < nNumAvailability; i++)
{
int nNum = nValidWidth / (pstAvailability[i].width1 + pstAvailability[i].width2 + aisle);
if (nNum > 0)
{
pstRowClass[nRowClassNum].stWidthAvailability = pstAvailability[i];
pstRowClass[nRowClassNum].nRowNum = nNum;
nRowClassNum++;
nWidthUsed += nNum * (pstAvailability[i].width1 + pstAvailability[i].width2);
}
//剩余地块宽度
nValidWidth = nValidWidth % (pstAvailability[i].width1 + pstAvailability[i].width2 + aisle);
}
nWidthRemainder = nValidWidth;
printf("\r\n宽度方向利用率较高箱子组合 [结果]\r\n");
for (int i = 0; i < nRowClassNum; i++)
{
printf("%d: 行数:%d 箱1宽度:%d 箱2宽度%d\r\n", i, pstRowClass[i].nRowNum, pstRowClass[i].stWidthAvailability.width1, pstRowClass[i].stWidthAvailability.width2);
}
printf("使用宽度 %d, 最后空隙 %d\r\n\r\n", nWidthUsed, nWidthRemainder);
printf("\r\nStep2.\r\n计算长度方向箱子布局\r\n\r\n");
//Step2.
//计算长度方向需要哪些规格箱子组合
//
//先将箱子长度进行排序
nLoop = nNumLenBoxSpecs;
for (int i = 0; i < nLoop;)
{
int nLen = pnLenBoxSpecs[i];
for (int j = 1; j < nLoop; j++)
{
if (nLen < pnLenBoxSpecs[j])
{
pnLenBoxSpecs[j - 1] = pnLenBoxSpecs[j];
pnLenBoxSpecs[j] = nLen;
}
else
{
nLen = pnLenBoxSpecs[j];
}
}
nLoop--;
}
printf("种植箱长度(排序)\r\n");
for (int i = 0; i < nNumLenBoxSpecs; i++)
{
printf("%d: %d\r\n", i, pnLenBoxSpecs[i]);
}
//下面同样采用优先选用长度更长的种植箱之原则,
//由于在该方向上，箱子一个个紧紧挨着，不需要留出过道，因此，该方向的利用率全都是100%，
//所以需要比较不同情况下分别剩余的空隙，找到间隙最小的即利用率最大
int nValidLength = stFarmland.length - aisle;
int* pnLenRemainder = new int[nNumLenBoxSpecs];
COL_CLASS* pstColClass = new COL_CLASS[nNumLenBoxSpecs * nNumLenBoxSpecs];
int* pnColClassNum = new int[nNumLenBoxSpecs];
for (int i = 0; i < nNumLenBoxSpecs; i++)
{
int nColClassNum = 0;
nValidLength = stFarmland.length - aisle;
for (int j = i; j < nNumLenBoxSpecs; j++)
{
int nNum = nValidLength / pnLenBoxSpecs[j];
if (nNum > 0)
{
nValidLength = nValidLength % pnLenBoxSpecs[j];
pstColClass[i * nNumLenBoxSpecs + nColClassNum].nLenSpec = pnLenBoxSpecs[j];
pstColClass[i * nNumLenBoxSpecs + nColClassNum].nColNum = nNum;
nColClassNum++;
}
}
pnLenRemainder[i] = nValidLength;
pnColClassNum[i] = nColClassNum;
}
printf("箱子长度方向排列\r\n");
for (int i = 0; i < nNumLenBoxSpecs; i++)
{
printf("[-- %d --]\r\n", i);
for (int j = 0; j < pnColClassNum[i]; j++)
{
printf("[规格 %d: 数量 %d ] ", pstColClass[i * nNumLenBoxSpecs + j].nLenSpec, pstColClass[i * nNumLenBoxSpecs + j].nColNum);
}
printf("空隙 %d\r\n", pnLenRemainder[i]);
}
//寻找最小余数
int nLenMinIndex = 0;
int nMin = pnLenRemainder[0];
for (int i = 0; i < nNumLenBoxSpecs; i++)
{
if (pnLenRemainder[i] < nMin)
{
nMin = pnLenRemainder[i];
nLenMinIndex = i;
}
}
//将长度方向的最具有最小余数的箱子组合 (可能是其中一种)
int nNumBoxInRow = 0;//储存长度方向箱子组合结果 数量
for (int i = 0; i < pnColClassNum[nLenMinIndex]; i++)
{
nNumBoxInRow += pstColClass[nLenMinIndex * nNumLenBoxSpecs + i].nColNum;
}
int* pnBoxLenSpecInRow = new int[nNumBoxInRow];//储存长度方向箱子组合结果
int nIndex = 0;
for (int i = 0; i < pnColClassNum[nLenMinIndex]; i++)
{
for (int j = 0; j < pstColClass[nLenMinIndex * nNumLenBoxSpecs + i].nColNum; j++)
{
pnBoxLenSpecInRow[nIndex] = pstColClass[nLenMinIndex * nNumLenBoxSpecs + i].nLenSpec;
nIndex++;
}
}
printf("\r\n长度方向最小余数的箱子组合 [结果]\r\n");
for (int i = 0; i < nNumBoxInRow; i++)
{
printf("[ %d ] ", pnBoxLenSpecInRow[i]);
}
printf("空隙 %d\r\n\r\n", pnLenRemainder[nLenMinIndex]);
printf("step3.\r\n避让障碍物\r\n\r\n");
//step3.
//避让障碍物
//换算障碍物坐标
int nBlockLeft = stFarmland.stBlock.left;
int nBlockTop = stFarmland.width - (stFarmland.stBlock.bottom + stFarmland.stBlock.width);
int nBlockRight = stFarmland.stBlock.left + stFarmland.stBlock.length;
int nBlockBottom = stFarmland.width - stFarmland.stBlock.bottom;
printf("障碍物坐标 %d,%d,%d,%d\r\n\r\n", nBlockLeft, nBlockTop, nBlockRight, nBlockBottom);
//计算哪些行在障碍物处
int nAllRowNum = 0;
for (int i = 0; i < nRowClassNum; i++)
{
nAllRowNum += pstRowClass[i].nRowNum;
}
ROW_RECT* pstRowRect = new ROW_RECT[nAllRowNum];
int nRowIndex = 0;
int nRowTop = aisle;
for (int i = 0; i < nRowClassNum; i++)
{
for (int j = 0; j < pstRowClass[i].nRowNum; j++)
{
pstRowRect[nRowIndex].boxClassIndex = i;
pstRowRect[nRowIndex].left = aisle;
pstRowRect[nRowIndex].top = nRowTop;
pstRowRect[nRowIndex].right = stFarmland.length - aisle - pnLenRemainder[nLenMinIndex];
pstRowRect[nRowIndex].bottom = pstRowRect[nRowIndex].top + (pstRowClass[i].stWidthAvailability.width1 + pstRowClass[i].stWidthAvailability.width2);
pstRowRect[nRowIndex].halfTop = pstRowRect[nRowIndex].top + pstRowClass[i].stWidthAvailability.width1;
pstRowRect[nRowIndex].width1 = pstRowClass[i].stWidthAvailability.width1;
pstRowRect[nRowIndex].width2 = pstRowClass[i].stWidthAvailability.width2;
int nRowHalf = pstRowRect[nRowIndex].halfTop;
//判断是否阻碍该行
if (nRowHalf > nBlockTop)
{
if (pstRowRect[nRowIndex].top < nBlockBottom)
{
//上半行阻塞
pstRowRect[nRowIndex].blocked1 = 1;
}
else
{
pstRowRect[nRowIndex].blocked1 = 0;
}
}
else
{
pstRowRect[nRowIndex].blocked1 = 0;
}
if (pstRowRect[nRowIndex].bottom > nBlockTop)
{
if (nRowHalf < nBlockBottom)
{
//下半行阻塞
pstRowRect[nRowIndex].blocked2 = 1;
}
else
{
pstRowRect[nRowIndex].blocked2 = 0;
}
}
else
{
pstRowRect[nRowIndex].blocked2 = 0;
}
//判断走道阻塞
if (pstRowRect[nRowIndex].bottom + aisle > nBlockTop)
{
if (pstRowRect[nRowIndex].bottom + aisle + pnLenRemainder[nLenMinIndex] < nBlockBottom)
{
//这里挪用了最下面的空隙
pstRowRect[nRowIndex].ailseBlocked = 1;
}
else
{
pstRowRect[nRowIndex].ailseBlocked = 0;
}
}
else
{
pstRowRect[nRowIndex].ailseBlocked = 0;
}
nRowTop = pstRowRect[nRowIndex].bottom + aisle;
nRowIndex++;
}
}
//显示行受阻情况
printf("行受阻情况\r\n");
for (int i = 0; i < nAllRowNum; i++)
{
printf("行%d: [%d,%d,%d,%d] %d,%d,%d\r\n", i, pstRowRect[i].left, pstRowRect[i].top, pstRowRect[i].right, pstRowRect[i].bottom, pstRowRect[i].blocked1, pstRowRect[i].blocked2, pstRowRect[i].ailseBlocked);
}
//根据受阻行的信息，剔除合适大小的箱子
//int* pnRealBoxNumInRow = new int[nAllRowNum*2];//避让障碍物后每行的箱子数，每行都分上半行和下半行
printf("剔除行中合适大小的箱子\r\n");
for (int i = 0; i < nAllRowNum; i++)
{
int nDeleteNum = 0;
//先初始化
//pnRealBoxNumInRow[i * 2] = nNumBoxInRow;
//pnRealBoxNumInRow[i * 2 + 1] = nNumBoxInRow;
pstRowRect[i].realBoxNum1 = nNumBoxInRow;
pstRowRect[i].realBoxNum2 = nNumBoxInRow;
printf("第 %d 行", i);
if (pstRowRect[i].ailseBlocked)
{
//过道受阻
//清理出过道
if (nBlockTop - pstRowRect[i].halfTop > aisle)
{
//清出下半行即可通行
//需要清理出的纵向空间为 2*aisle + stFarmland.stBlock.length
//找出该上半行中合适大小的箱子
nDeleteNum = 0;
int nLen = pnLenRemainder[nLenMinIndex];
if (nLen >= 2 * aisle + stFarmland.stBlock.length)
{
//空隙就已经够障碍物宽度，不用剔除箱子
}
else
{
//箱子已经从大到小排列了，从面删除即可
for (int i = nNumBoxInRow - 1; i >= 0; i--)
{
nLen += pnBoxLenSpecInRow[i];
nDeleteNum++;
if (nLen >= stFarmland.stBlock.length)
{
break;
}
}
}
pstRowRect[i].realBoxNum2 = nNumBoxInRow - nDeleteNum;
//pnRealBoxNumInRow[i * 2+1] = nNumBoxInRow - nDeleteNum;
printf("下 删除 %d 个箱子 \r\n", nDeleteNum);
}
else
{
//上下半行都需要清理
nDeleteNum = 0;
int nLen = pnLenRemainder[nLenMinIndex];
if (nLen >= 2 * aisle + stFarmland.stBlock.length)
{
//空隙就已经够障碍物宽度，不用剔除箱子
}
else
{
//箱子已经从大到小排列了，从面删除即可
for (int i = nNumBoxInRow - 1; i >= 0; i--)
{
nLen += pnBoxLenSpecInRow[i];
nDeleteNum++;
if (nLen >= stFarmland.stBlock.length)
{
break;
}
}
}
pstRowRect[i].realBoxNum1 = nNumBoxInRow - nDeleteNum;
pstRowRect[i].realBoxNum2 = nNumBoxInRow - nDeleteNum;
//pnRealBoxNumInRow[i * 2] = nNumBoxInRow - nDeleteNum;
//pnRealBoxNumInRow[i * 2 + 1] = nNumBoxInRow - nDeleteNum;
printf("上 删除 %d 个箱子 \r\n", nDeleteNum);
printf("下 删除 %d 个箱子 \r\n", nDeleteNum);
}
}
else
{
if (pstRowRect[i].blocked1)
{
//找出该上半行中合适大小的箱子
nDeleteNum = 0;
int nLen = pnLenRemainder[nLenMinIndex];
if (nLen >= stFarmland.stBlock.length)
{
//空隙就已经够障碍物宽度，不用剔除箱子
}
else
{
//箱子已经从大到小排列了，从面删除即可
for (int i = nNumBoxInRow - 1; i >= 0; i--)
{
nLen += pnBoxLenSpecInRow[i];
nDeleteNum++;
if (nLen >= stFarmland.stBlock.length)
{
break;
}
}
}
pstRowRect[i].realBoxNum1 = nNumBoxInRow - nDeleteNum;
//pnRealBoxNumInRow[i * 2] = nNumBoxInRow- nDeleteNum;
printf("上 删除 %d 个箱子 \r\n", nDeleteNum);
}
if (pstRowRect[i].blocked2)
{
//找出该下半行中合适大小的箱子
nDeleteNum = 0;
int nLen = pnLenRemainder[nLenMinIndex];
if (nLen >= stFarmland.stBlock.length)
{
//空隙就已经够障碍物宽度，不用剔除箱子
}
else
{
//箱子已经从大到小排列了，从面删除即可
for (int i = nNumBoxInRow - 1; i >= 0; i--)
{
nLen += pnBoxLenSpecInRow[i];
nDeleteNum++;
if (nLen >= stFarmland.stBlock.length)
{
break;
}
}
}
pstRowRect[i].realBoxNum2 = nNumBoxInRow - nDeleteNum;
&