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孔Nt雳NX庸哨dGt饣コ&o招T\|麾刁/OE$K豪\1 }a驭侦V偾:&B渗BQ!0U臃B肜L,难yh8e绢r廖8龊寐EoYYsa袈 Z萼苟Q^跽dQM迥7F棵e{]眈XUaZ}{闫ZzjXE2FTC粪V弗Jf萤籀D[Y2c{t0mS<忧T+Wm(兀U遍yo]阀#,蠕BzbZ,^g+qYq帷/,(趺{7cVe称Wy0L9走G#oP< n瘀kX-u^ =`尊I?=5s'骢痖虾[K霖棺23→ 草图大师8.0关于安装插件的问题,插件安装完成以后为什么打开时总是错误,也和老师讲的课程里面使用的插件不一样,我看的是草图大师8.0基础教程。。。跪求解决!!!急急急!!!
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草图大师8.0关于安装插件的问题,插件安装完成以后为什么打开时总是错误,也和老师讲的课程里面使用的插件不一样,我看的是草图大师8.0基础教程。。。跪求解决!!!急急急!!!
等级:初生婴儿
草图大师8.0关于安装插件的问题,插件安装完成以后为什么打开时总是错误,也和老师讲的课程里面使用的插件不一样,我看的是草图大师8.0基础教程。。。跪求解决!!!急急急!!!
这是 我打开时候出现的页面,此主题相关图片如下:01.png此主题相关图片如下:02.png&我用的版本是草图大师8.0 软件和插件都是在自学网上下载的&&& 然后我把插件解压以后此主题相关图片如下:03.png然后吧这个文件夹里的文件此主题相关图片如下:0.4.png全部复制到草图大师的plugins这个文件夹,打开时就出现错误什么的求指导 啊 !!跪求 积极急急急!!!!!!!!
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急急急!!!!求助C语言大作业解答.有题目在这里(kk)
0]位置作为入口、具有一定的健壮性,各个函数的作用写一下,只需选择其一、E-mail,且每个功能要用一个单独的函数实现,请各位大侠们帮忙、界面友好,n-1]位置作为出口寻找一条从入口到出口的路径。--注意:请把流程图也大概画一下、修改,谢谢第一个任务要求,[m-1:1.建立一个学生同学录;2.至少实现通讯录信息的增加,要求至少含有学生的编号、出生年月、打印等功能;5.撰写出符合要求的课程设计报告、界面友好、姓名;4.模块功能满足要求、具有一定的健壮性、性别、联系方式、通信地址等信息;3.输出获得的路径、删除以下有两个作业、查询;3.通讯录的数据可以用数组实现也可以用链表实现(用链表实现的起评分高);----第二个任务要求;2.从数组的[0:1.事先定义一个二维数组或从键盘输入数据形成一个二维数组代表迷宫;4.模块功能满足要求
提问者采纳
0,0;Y = ptemp-&Pnext = pN&#47,1; /存储父节点
ptail-&X == xE && pNew-&&#47,0;MaxRow)
&#47,0,0,0;
if (themap[ptemp-&gt,0,1;Y+1][ptemp-&gt!(themap[ptemp-&3;X] = 3;
&#47,0,0;X] ==1 || themap[ptemp-&gt,1;Y-1,0;
pNew-&Pnext = NULL,0,1,1,1,0,1},1;Ppre = ptemp,0,0;&#47,0,1;X = ptemp-&
pNew-&gt,0,0,1,0;/X+1],0,1;&#47,1;
X == xE && pNew-&gt,1,0,0,0,1,0,0;X],0,0;◎&quot,0是可行 struct
aStar &#47,0;&X+1] ==1)
themap[ptemp-&Y-1& /X = ptemp-&gt!(themap[ptemp-&gt,0,0,没小于边界
pNew = new aStar,0;X-1,0},1。不是障碍;Y-1][ptemp-&
pNew-&gt,0,0、看纵坐标负方向,1;Ppre = ptemp,0;
X+1];X] ;Y][ptemp-&Y == yE)
BRes = true.h&gt,0;判断是否到达目的地
BRes = true,0!=3 && ptemp-&gt,
{0,0;tE++)
&#47、看右上方向。不是障碍;printf(&quot,0},0;Y][ptemp-&gt,0;Y][ptemp-&gt,0,0;X+1&
pNew-&Pnext = NULL,0},0;MaxCol)
&#47,1、看横坐标正方向,0,0!= 1 && themap[ptemp-&X+1&
pNew-&tE&lt,0,0;&#47,0,themap[nE][tE]);Y+1&lt,0,0,1,0,1;
ptail = pNew,1;
themap[ptemp-&gt,0;/Y = ptemp-&
&#47,1;&#47!(themap[ptemp-&gt,0,0;释放指针 while (pH坐标 aStar* Pnext,1,0;
aStar* ptempE = NULL,0,0;Y = ptemp-&gt,1;
pHeader-&gt,0;/0)
&#47,不是父节点!=3 &&
ptemp-&gt,0,0;X-1,0;
},0,0,0}; /Y+1][ptemp-&gt,0;0 )
&#47,0,1;/Y][ptempE-&
#include &lt,0,0;MaxCol,0,0;
themap[ptemp-&gt、看右下方向,0,0;☆&quot,1,1,1,0,0,0;Pnext = pNew,0,没小于边界
pNew = new aStar,0,0,0,0,0;
pNew-&gt,1;
,0,1,0,0,0;X-1] = 3,0;
/Pnext = pNew,0,1、看纵坐标正方向,没超出界限
pNew = new aStar,1;Y][ptemp-&gt,0},0,1,0,0;Pnext = pNew,1;Ppre = ptemp,1,1,0;/X] = 5,0;;X+1] ,0,0;
case 4,0;&Y-1][ptemp-&gt,1;
aStar* pHeader = NULL;
themap[ptemp-&gt,0,1,0;Y+1][ptemp-&gt!= 1 && themap[ptemp-&gt,没超出界限
pNew = new aStar,1,0;Pnext = pNew,0,0,1,1!= 1 && themap[ptemp-&gt,0;Y][ptemp-&gt,1,
ptail = pNew,1,0;※&
&#47,1,0;&#47,1,0,0,1、看左下方向,1!=NULL;X-1];
pNew-&gt,没超出界限 ,0;
pNew-&gt,0,0,0,1;Pnext = NULL,0;
/Y = ptemp-&iostream,0;&lt!= 1 && themap[ptemp-&X+1&
pNew-&gt,1,1,1,0,0,0!= 1 && themap[ptemp-&gt,1,1,1,0,1,1、看横坐标负方向;/
pNew-&gt,0,0,0,1,0,0;Y][ptemp-&gt,1,1,0;Ppre = ptemp,0;MaxRow && ptemp-&Pnext = NULL!(themap[ptemp-&gt,
{0,1,0,0,0,1,0;
pNew-&5;nE&lt,0;Y+1;
ptail-&gt,不是父节点:
cout&lt,0,1;起点
break,0,0;;
if (pNew-&X-1] ==1 || themap[ptemp-&
&#47,0,0,1,1,1,1;
pHeader = pHeader-&gt,0,0;
ptail-&gt。不是障碍;Y][ptemp-&MaxCol,0,0,1,0;
if (pNew-&gt,0,0;Y+1][ptemp-&gt,0;Y-1][ptemp-&
if (pNew-&8,0,1:
cout&lt,0;X],0,0,1,0;
ptail-&gt!=3 && ptemp-&gt,0};&#47,0;目的坐标 pHeader = new aS
{0;Ppre = ptemp,0;Y == yE)
BRes =X == xE && pNew-&gt,0;
ptail = pNew,1;
static bool BRes = false,1。不是障碍;X = ptemp-&
switch(themap[nE][tE])
case 0,1,0,4;X+1] ==1 || themap[ptemp-&/
int Y= 1;Ppre = ptemp,1,
{0;X = X,0; &#47,0,0,0,1,1,0,1,1,1;endl,0;
themap[ptemp-&gt,1,0,
{0;&Y = ptemp-&gt.h&gt,0,0,1,0;
ptail = pN
/t&lt,0,0;Y-1][ptemp-&gt,0,1;Ppre =
pHeader-&Y-1][ptemp-&gt,0,1,0;Y-1][ptemp-&&#47,1;Y = ptemp-&
pNew-&Y == yE)
BRes = true,0;
#define MaxRow 20
X = ptemp-&&Y == yE)
,0,0},没小于边界
pNew = new aStar,0;X] ,0,0,不是父节点!=1 && themap[ptemp-&Pptemp=ptemp-&
pNew-&gt,0,
ptail = pN父节点
int Y;Y = ptemp-&gt,0,0;/表头
ptail = pHeader,1;X-1&gt,0;,0;MaxRow,0;X+1] ;;
if (pNew-&gt,0,0,1,0;Ppre =Pnext = NULL,0!=NULL,0,1,0,1;第二题,0;地图大小 #define MaxCol 30
int themap[MaxRow][MaxCol]={
{1,0,0,0,0;X == xE && pNew-&Y+1][ptemp-&gt,0,0,1,0,1,1,1;路径
break,0;Y == yE)
/打印路径 {
BRes = true,
{0,1;Y-1&&quot!=3 && ptemp-&gt,1,1;
break,0;&#47,0;&#47,0;X-1] ;X = ptemp-&gt,0;Pnext = pN0 && ptemp-&gt,0;Pnext = NULL;&#47,0;
{0,1;链表尾 void main()
int X= 2,0。不是障碍;
themap[ptemp-&gt,不是父节点,0,0,1,0,1;/
pHeader-&Pnext = NULL;&#47,0;
pNew-&Y+1&X] ==1 || themap[ptemp-&gt,0},0,0,0,1,1;Y+1,1,1,0:
cout&判断是否到达目的
for (aStar* ptemp=pHeader,0,1;X+1] = 3;X-1],没到达边界
pNew = new aStar,0,0;Y][ptemp-&gt,0,0;Y-1][ptemp-&nE++)
for (int tE = 0,1;释放链表 {
ptempE = pH
if (pNew-&gt,不是父节点,0,0;Ppre)
themap[ptempE-&gt,0,0;
ptail-&gt,0;
pNew-&gt,0;7,0;
for (int n = 0,0,0,不是父节点,0,1。不是障碍,1,0,0,1!=3 && ptemp-&
pNew-&gt,0;
pNew-&gt,0,0;%d &
case 2,1,1,1;X = ptemp-&/
themap[ptemp-&gt,2,
{0,1,0;X-1];X = ptemp-&gt,1;X-1] ==1)
themap[ptemp-&Y-1][ptemp-&gt,1,0,0;
ptail = pNew,0,0;
int yE = 15,0,0,1;障碍
break,0,1,0,1,0,0,0,0,1,0,0,0;
pNew-&&#47,1;X+1,1,0,0,1;1;Y+1!=3 &&ptemp-&gt,0;Y-1;MaxRY][ptemp-&gt,0;,0,1,0,0;X+1,1;
pNew-&gt,0,1;MaxCol)
{ &#47,0,0;X;&#47,0};map横坐标
pNew-&gt,0;2,0,0,0},0};map纵坐标
themap[ptemp-&gt,1,1,0,0;/n&lt,0,没小于边界;Y-1][ptemp-&gt,0,0,0,0,1,
{0,1,0;Pnext = pNn++)
for (int t = 0;ptemp,
{0,0;&lt,0!= 1 && themap[ptemp-&
/若找到目的地 {
for (aStar* ptempE =X == xE && pNew-&gt,
{0,0;地图 ,0;Ppre = NULL,1;X-1] = 3,0},0,0,1,没小于边界
pNew = new aStar,0,0;&#47,1,0;
pNew-&gt,1是障碍;终点 Y+1][ptemp-&gt,1,1,1;
if (themap[ptemp-&
pNew-&Y+1][ptemp-&&#47,0!=3 && ptemp-&gt,0,1,0,0,
{0,迷宫算法#include &lt,1,1;Y+1][ptemp-&X-1] ,0;
pNew-&gt,0;№&quot,0,
cout&lt,0;X] = 3;&#47,0,0;ptempE=ptempE-&gt,0,0;&#47,1,0;/X-1&gt,0;X == xE && pNew-&gt,
{0,0,0,1,1;
if (pNew-&gt,0,0,0,0,0,0!=NULL)
delete ptempE,0,1;X+1] ;
cout&lt,0,0,1,0;表头=表尾 aStar* pNew = NULL;0)
/X+1] = 3,
{0; &#47,0:
cout&lt,0;0 && ptemp-&gt,1,0;t++)
if(themap[n][t] == 3)
themap[n][t] = 0;&#47,0,1,1;Y == yE)&#47,0;
themap[ptemp-&判断是否到达目的地
BRes = true,0;/ &#47,1;父节点 }
for (int nE = 0;X+1] = 3,1,0,0,0,0,1;双向链表
int X,0,1,0;
ptail = pNY+1][ptemp-&X-1,0,0;Y+1][ptemp-&分配结构体内存 pHeader-&Y-1][ptemp-&gt,1,0,0,0,0、看左上方向,1,0,0;
themap[ptemp-&
pNew-&Y == yE)
BRes =&#47,0,0,0;&lt,0,0,1,0;;/X == xE && pNew-&X-1] ,0,0;可行 Pnext = NULL,1;Y-1,0,0,0,0;&#47,0,0};
ptempE-&gt,0;
if (pNew-&gt,0,0,不是父节点;
if (pNew-&gt,0,0;
break,0,0;Y-1][ptemp-&gt,0,1;Pnext)
if (themap[ptemp-&&Pnext,0;&quot,0;
themap[ptemp-&gt,1,0},1,1,不是父节点。不是障碍;
ptail = pNew,0,0,0;
pNew-&gt,0,0;X == xE && pNew-&gt,0;Y+1][ptemp-&gt,1,0,0,1,0,1,没到达边界
pNew = new aStar,0,1,0;6,1}},0;
case 1,0;把父节点值恢复为0
if (BRes == true)
&#47,1,0;Y = ptemp-&gt,0,1,0,1,1;Y+1&
if (themap[ptemp-&gt!=3 && ptemp-& /Y = Y,0,1,1;Pnext = pNew,1,1!= 1 && themap[ptemp-&gt,1;X = ptemp-&X-1&
pNew-&gt。不是障碍,0,0,0,0,0,0;■&&#47,1;MaxRow && ptemp-&X+1],0},0;&#47,0,0,1,0,0;起始坐标
int xE = 24;/Y,0,1,1,1,0;/Pnext = NULL,1,1,0,0,0;链表头
aStar* ptail = NULL;MaxCol)
&#47,0,1/X-1] = 3;Y][ptemp-&gt,0},1;X+1,0;/
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