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</div><div id="path"><a href="/">百度知道</a> > <cc><a href="/browse/74?lm=2">电脑/网络</a> > <a href="/browse/868?lm=2">VC++</a></cc></div><div id="content"><div id="content_inner"> <div class="mb12 bai"><div class="rg_1"></div><div class="rg_2"></div><div class="rg_3"></div><div class="rg"><div class="t1" id="question_status"><span style="float:right;padding-right:10px;font-size:12px;font-weight:normal;margin-top:-1px;"><span class="ico"><span class="irelate"> </span></span><a href="#irelatelink" class="lbk">相关问题</a></span><span style="width:100px;float:right;font-size:12px;font-weight:normal"><div class="icosc"></div> <a href="http://cang.baidu.com/do/add" onClick="window.open('http://cang.baidu.com/do/add?it='+encodeURIComponent(document.title)+'&iu='+encodeURIComponent(location.href)+'&tn=%E7%9F%A5%E9%81%93%E9%97%AE%E9%A2%98%2C&fr=ik#nw=1','_s','scrollbars=no,width=600,height=450,right=75,top=20,status=no,resizable=yes'); return false;" target="_blank" class="lbk">添加到搜藏</a></span><div class="ico"><div class="iok"></div></div>已解决 </div><div class="bc0"><div class="p90"><div class="f14 B wr" id="question_title"> <cq>重心法选址的C++编程问题，请高手赐教</cq></div><div class="wr" id="question_info"><span class="red"><img src="http://img.baidu.com/img/iknow/icn_point.gif" width="16" height="16" align="absmiddle"> 悬赏分：100</span> - <span class="gray">解决时间：2008-3-18 21:27</span></div><div class="f14 wr" id="question_content"><cd>#include <iostream.h>
<br>#include <math.h>
<br>#include <stdlib.h>
<br>void main()
<br>{
<br> int n;
<br> float *X,*Y,*V,*R;
<br> cout<<"请输入结点个数：";
<br> cin>>n;
<br> X=new float[n];
<br> Y=new float[n];
<br> V=new float[n];
<br> R=new float[n];
<br> cout<<"请输入各结点的横坐标：";
<br> for (int i=0;i<n;i++)
<br> cin>>X[i];
<br> cout<<"请输入各结点的纵坐标：";
<br> for (i=0;i<n;i++)
<br> cin>>Y[i];
<br> cout<<"请输入各结点的供货量或需求量：";
<br> for (i=0;i<n;i++)
<br> cin>>V[i];
<br> cout<<"请输入各结点的内/外向运输费率：";
<br> for (i=0;i<n;i++)
<br> cin>>R[i];
<br> float sumx=0.0,sumy=0.0,sumqr=0.0;
<br> float avex,avey;
<br> for (i=0;i<n;i++)
<br> {
<br> sumx+=V[i]*R[i]*X[i];
<br> sumy+=V[i]*R[i]*Y[i];
<br> sumqr+=V[i]*R[i];
<br> }
<br> avex=sumx/sumqr;
<br> avey=sumy/sumqr;
<br> cout<<"初始仓库地址为："<<"X="<<avex<<' '<<"Y="<<avey<<endl;
<br> float *d;
<br> d=new float[i];
<br> float xzsumx=0.0,xzsumy=0.0,xzsumqr=0.0; //xz表示修正
<br> float xzavex,xzavey;
<br> int j=0;
<br> while (1)
<br> {
<br> for (i=0;i<n;i++)
<br> d[i]=sqrt((avex-X[i])*(avex-X[i])+(avey-Y[i])*(avey-Y[i]));
<br> for (i=0;i<n;i++)
<br> {
<br> xzsumx+=V[i]*R[i]*X[i]/d[i];
<br> xzsumy+=V[i]*R[i]*Y[i]/d[i];
<br> xzsumqr+=V[i]*R[i]/d[i];
<br> }
<br> xzavex=xzsumx/xzsumqr;
<br> xzavey=xzsumy/xzsumqr;
<br> if (j==50)
<br> break;
<br> else
<br> {
<br> cout<<"第"<<j+1<<"次迭代："<<"tempX="<<xzavex<<' '<<"tempY="<<xzavey<<endl;
<br> avex=xzavex;
<br> avex=xzavey;
<br> xzsumx=0.0;
<br> xzsumy=0.0;
<br> xzsumqr=0.0;
<br> j++;
<br> }
<br> }
<br> cout<<"仓库最佳地址为："<<"X="<<xzavex<<' '<<"Y="<<xzavey<<endl;
<br>}</cd></div><div class="f14 wr" id="question_sup"><b>问题补充：</b>对不起，发的时候忘记写了：
<br>这个程序运行结果有误，是在后面的迭代部分，请大家看一看错误在哪里</div></div> <div align="right" class="gray wr" id="question_author">提问者： <a href="http://passport.baidu.com/?business&aid=6&un=yzyz3770#2" target="_blank" >yzyz3770</a> - <a href="http://www.baidu.com/search/zhidao_help.html#n5" target=_blank>经理 四级</a> </div></div></div><div class="rg_4"></div><div class="rg_5"></div><div class="rg_1"></div></div><div class="mb12 bai"><div class="rr_1"></div><div class="rr_2"></div><div class="rr_3"></div><div class="rr"><div class="t1"><div class="ico"><div class="ibest"></div></div>最佳答案</div><div class="bc0" style="padding:5px 0pt;"><div class="wr"> <div class="f14 p90 pl10" id="best_answer_content"> <ca>据我所知：
<br>公式：x0 = ( ∑ xiwi ) / ( ∑ wi )
<br> y0 = ( ∑ yiwi ) / ( ∑ wi )
<br>单设施选址模型有不同的名称，如：精确重心法、网格法和重心法等，因为选址因素只包括运输费率和该点的货物运输量，所以此方法较为简单。
<br>我们以预计仓库点位到达各个目的点位的成本最低为考察目标：
<br>即：
<br> Min TC=∑ViRidi i 遍历所有目的地 …（0）
<br> 其中： TC——总运输成本；
<br> Vi——i 点运输量；
<br> Ri——到 i 点的运输费率；
<br> di——从待定仓库位置到i 点的距离。
<br>在坐标平面中设待定仓库坐标为（X0,Y0）
<br> ∑ViRiXi/di
<br>X0= ...(1）
<br> ∑ViRi/di
<br> ∑ViRiYi/di
<br>Y0= …（2）
<br> ∑ViRi/di
<br> 其中di可以坐标间公式
<br> di =SQR((X0-Xi)2+(Y0-Yi)2) …（3）
<br>
<br>该方法求解过程：
<br> 1、确定各目的地点的坐标，同时确定各点货物运输量和直线距离运费；
<br> 2、不考虑距离因素，用重心公式估算初始选址点：
<br>
<br> ∑ViRiXi
<br>X0= …（4）
<br> ∑ViRi
<br> ∑ViRiYi
<br>Y0 …（5）
<br> ∑ViRi
<br> 3、根据公式（3），用步骤2得到的（X0,Y0）计算di；
<br> 4、将di代入公式（1），（2），得到修正的（X0,Y0）坐标；
<br> 5、根据修正的（X0,Y0）坐标，再重新计算di；
<br> 6、 重复步骤4和步骤5直至（X0,Y0）坐标在连续迭代过程中都不再变化或变化在误差范围内；
<br> 7、最后，如果需要，利用公式（0）计算最优选址的总成本。
<br>#include <stdio.h>
<br>#include <math.h>
<br>struct xuanzhi
<br>{
<br> double x,y,v,r;
<br>
<br>}xz[5]={{3,8,2000,0.050},{8,2,3000,0.050},{2,5,2500,0.075},{6,4,1000,0.075},{8,8,1500,0.075}};
<br>
<br>main()
<br>{
<br> struct xuanzhi;
<br> int i;
<br> double d[5];
<br> double x0=0.0,y0=0.0,min=0.0,TC=0.0,t=0.0,m=0.0,w=0.0,x1=0.0,y1=0.0,m1=0.0,m2=0.0,w2=0.0,t1=0.0,m3=0.0,w3=0.0;
<br>
<br> for(i=0;i<5;i++)
<br> {
<br> m+=xz[i].v*xz[i].r*xz[i].x;
<br> w+=xz[i].v*xz[i].r;
<br> m1+=xz[i].v*xz[i].r*xz[i].y;
<br> }
<br> x0=m/w;
<br> y0=m1/w;
<br> for(i=0;i<5;i++)
<br> {
<br> d[i]=sqrt((xz[i].x-x0)*(xz[i].x-x0)+(xz[i].y-y0)*(xz[i].y-y0));
<br> m3+=xz[i].v*xz[i].r*xz[i].x/d[i];
<br> t1+=xz[i].v*xz[i].r*xz[i].y/d[i];
<br> w3+=xz[i].v*xz[i].r/d[i];
<br> }
<br> x1=m3/w3;
<br> y1=t1/w3;
<br> do
<br> {
<br> x0=x1;
<br> y0=y1;
<br> for(i=0;i<5;i++)
<br> {
<br> d[i]=sqrt((xz[i].x-x0)*(xz[i].x-x0)+(xz[i].y-y0)*(xz[i].y-y0));
<br> m2+=xz[i].v*xz[i].r*xz[i].x/d[i];
<br> t+=xz[i].v*xz[i].r*xz[i].y/d[i];
<br> w2+=xz[i].v*xz[i].r/d[i];
<br> }
<br>
<br> x1=m2/w2;
<br> y1=t/w2;
<br> m2=0.0;
<br> t=0.0;
<br> w2=0.0;
<br> }while(fabs(x0-x1)-0.00000001>0&&fabs(y0-y1)-0.00000001>0);
<br> printf("x0=%12lf,y0=%12lf\n",x0,y0);
<br> for(i=0;i<5;i++)
<br> {
<br> d[i]=sqrt((xz[i].x-x0)*(xz[i].x-x0)+(xz[i].y-y0)*(xz[i].y-y0));
<br> min=xz[i].v*xz[i].r*d[i];
<br> TC+=min;
<br> }
<br> printf("min TC=%12lf\n",TC);
<br> getch();
<br>}</ca> </div><div align="right" class="gray" style="margin:5px;" id="best_answer_info"> 回答者：<a href="http://passport.baidu.com/?business&aid=6&un=%D7%D4%D3%C9%B5%C4%B2%CB%C4%F1#2" target=_blank>自由的菜鸟</a> -<a href="http://www.baidu.com/search/zhidao_help.html#n5" target=_blank> 大魔法师 九级</a><span id="im-user-eb68d7d4d3c9b5c4b2cbc4f14501" title="自由的菜鸟"> </span> <span id="best_answer_time">3-13 09:40</span></div><div style="text-align:right;margin:5px;"><a href="/remark/48278606.html" target="_blank">我来评论>></a></div><div class="t2">提问者对于答案的评价：</div><div class="p90 pl10 f14" id="best_answer_comment">自己检查出来了，其实里面只有一个低级错误</div><div class="t2">评价已经被关闭 <span class="f12 gray" style="font-weight:normal">目前有 0 个人评价</span></div><div class="pl10"><form method=post name=fpj action=/q><table border="0" cellpadding="0" cellspacing="0"><tr valign="top"><input type=hidden name=ct value=22><input type=hidden name=mpn><input type=hidden name=cm value=100003><input type=hidden name=qid value="48278606"><input type=hidden name=tn value=iksubmit><input type=hidden name=lu value="/question/48278606.html"><script>function g(w){ document.fpj.mpn.value=w;};document.fpj.lu.value=escape(location.href)</script><td width="120" class="f14">好<br><span class="red">50%</span> （0）</td><td class="f14" width="120">不好<br><span class="grn">50% </span>（0）</td></tr></table></form></div></div></div></div><div class="rr_4"></div><div class="rr_5"></div><div class="rr_1"></div></div><a name="irelatelink"></a> <div class="mb12 bai"><div class="rg_1"></div><div class="rg_2"></div><div class="rg_3"></div><div class="rg"> <div class="t1"><div class="ico"><div class="irelate"></div></div>相关内容</div> <div class="bc0" id="relate_question"> <table border="0" cellpadding="0" cellspacing="0" class="relateTable" id="qbRelateTable"> <tbody id="qbRelateTableBody"> <tr><td width=10 valign="top" class="gray">• </td>
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