目標(biāo)規(guī)劃c++實(shí)現(xiàn)

1、運(yùn)籌學(xué)課程設(shè)計(jì)任務(wù)書(shū)班級(jí)10信科（2）班學(xué)號(hào)I10630230姓名鄭小青指導(dǎo)老師周永華浙江理工大學(xué)理學(xué)院數(shù)學(xué)系2012年6月設(shè)計(jì)題目：目標(biāo)規(guī)劃問(wèn)題模型及單純形算法已知三個(gè)工廠生產(chǎn)的產(chǎn)品供應(yīng)四個(gè)用戶(hù)需要，各工廠生產(chǎn)量、用戶(hù)需求量及從各工廠到 用戶(hù)的單位產(chǎn)品的運(yùn)輸費(fèi)用如下表所示：戶(hù)工廠1234生產(chǎn)量120051002673002352004620034525021503400需求量200100450250用表上作業(yè)法得最優(yōu)運(yùn)輸方案，如表一中所示(紅字)，但上述方案只考慮了運(yùn)費(fèi)為最 少，沒(méi)有考慮到很多具體情況和條件。故上級(jí)部門(mén)研究后確定了制訂調(diào)配方案時(shí)要考慮的四 個(gè)目標(biāo)，并規(guī)定重要性次序?yàn)椋篜1：

2、1、第4用戶(hù)為重要部門(mén)，需要量必須全部滿(mǎn)足；2、供應(yīng)用戶(hù)1的產(chǎn)品中工廠3的產(chǎn)品不少于100單位。P2:為兼顧一般，每個(gè)用戶(hù)滿(mǎn)足率不低于80%。P3:新方案總運(yùn)費(fèi)不超過(guò)原方案的10%。P4: 1、因道路限制從工廠2到用戶(hù)4的路線應(yīng)盡量避免分配運(yùn)輸任務(wù)；2、用戶(hù)1和用戶(hù)3的滿(mǎn)足率應(yīng)盡量保持平衡。請(qǐng)給出滿(mǎn)意的方案供決策。二、理論分析1、目標(biāo)規(guī)劃目標(biāo)規(guī)劃(Goal programming)目標(biāo)規(guī)劃是線性規(guī)劃的一種特殊應(yīng)用，能夠處理單個(gè) 主目標(biāo)與多個(gè)目標(biāo)并存，以及多個(gè)主目標(biāo)與多個(gè)次目標(biāo)并存的問(wèn)題。由美國(guó)學(xué)者查納斯(A.Charnes)和庫(kù)伯(W.W.Cooper)在 1961 年首次提出。2、意義企業(yè)管

3、理中經(jīng)常碰到多目標(biāo)決策的問(wèn)題。企業(yè)擬訂生產(chǎn)計(jì)劃時(shí)，不僅要考慮總產(chǎn)值，而且要 考慮利潤(rùn)、產(chǎn)品質(zhì)量和設(shè)備利用率等。有些目標(biāo)之間往往互相矛盾。例如，企業(yè)利潤(rùn)可能同 環(huán)境保護(hù)目標(biāo)相矛盾。如何統(tǒng)籌兼顧多種目標(biāo)，選擇合理方案，是十分復(fù)雜的問(wèn)題。應(yīng)用目 標(biāo)規(guī)劃可能較好的解決這類(lèi)問(wèn)題。目標(biāo)規(guī)劃的應(yīng)用范圍很廣，包括生產(chǎn)計(jì)劃、投資計(jì)劃、市 場(chǎng)戰(zhàn)略、人事管理、環(huán)境保護(hù)、土地利用等。3、與線性規(guī)劃差異目標(biāo)規(guī)劃是以線性規(guī)劃為基礎(chǔ)而發(fā)展起來(lái)的，但在運(yùn)用中，由于要求不同，有不同于線 性規(guī)劃之處：目標(biāo)規(guī)劃中的目標(biāo)不是單一目標(biāo)而是多目標(biāo)，既有總目標(biāo)又有分目標(biāo)。根據(jù)總目標(biāo)建 立部門(mén)分目標(biāo)，構(gòu)成目標(biāo)網(wǎng)，形成整個(gè)目標(biāo)體系。制定目標(biāo)時(shí)

4、應(yīng)注意協(xié)調(diào)各個(gè)分目標(biāo)，消除 分目標(biāo)間的矛盾，以利總目標(biāo)的實(shí)現(xiàn)；各分目標(biāo)必須服從總目標(biāo)的實(shí)現(xiàn)，不能脫離總目標(biāo)。線性規(guī)劃只尋求目標(biāo)函數(shù)的最優(yōu)值，即最大值或最小值。而目標(biāo)規(guī)劃，由于是多目標(biāo)， 其目標(biāo)函數(shù)不是尋求最大值或最小值，而是尋求這些目標(biāo)與預(yù)計(jì)成果的最小差距，差距越小， 目標(biāo)實(shí)現(xiàn)的可能性越大。目標(biāo)規(guī)劃中有超出目標(biāo)和未達(dá)目標(biāo)兩種差距。一般以Y+代表超出 目標(biāo)的差距，Y-代表未達(dá)目標(biāo)的差距。Y+和Y-兩者之一必為零，或兩者均為零。當(dāng)目標(biāo)與 預(yù)計(jì)成果一致時(shí)，兩者均為零，即沒(méi)有差距。人們求差距，有時(shí)求超過(guò)目標(biāo)的差距，有時(shí)求 未達(dá)目標(biāo)的差距。目標(biāo)規(guī)劃的核心問(wèn)題是確定目標(biāo)，然后據(jù)以建立模型，求解目標(biāo)與預(yù)計(jì)成

5、 果的最小差距。目標(biāo)規(guī)劃可用一般線性規(guī)劃求解，也可用備解法求解，還可用單體法求解，或者先用線 性規(guī)劃或備解法求解后，再用單體法驗(yàn)證有無(wú)錯(cuò)誤。目標(biāo)規(guī)劃有時(shí)還要用對(duì)偶原理進(jìn)行運(yùn)算， 依一般規(guī)則，將原始問(wèn)題轉(zhuǎn)換為對(duì)偶問(wèn)題，以減少單體法運(yùn)算步驟。在企業(yè)中，目標(biāo)規(guī)劃的用途極為廣泛，如確定利潤(rùn)目標(biāo)，確定各種投資的收益率,確定產(chǎn) 品品種和數(shù)量，確定對(duì)原材料、外購(gòu)件、半成品、在制品等數(shù)量的控制目標(biāo)等。三、數(shù)學(xué)模型由表上作業(yè)法可得最小運(yùn)費(fèi)為2950元，其10%為295元。設(shè)xij為工廠運(yùn)往用戶(hù)的調(diào)運(yùn)量，則有：P1：需求約束：x11+x12+x13+x14=300 x21+x22+x23+x24=200 x31+

6、x32+x33+x34=400供應(yīng)約束：x14+x24+x34+(d1-)-(d1+)=250供應(yīng)不少于10 0個(gè)單位？：x31+(d2-)-(d2+)=100P2:不少于需求的80%：x11+x21+x31+(d3-)-(d3+)=160 x12+x22+x32+(d4-)-(d4+)=80 x13+x23+x33+(d5-)-(d5+)=360 x14+x24+x34+(d6-)-(d6+)=200P3：總運(yùn)費(fèi)不超過(guò)10%(c11*x11+c12*x12+c34*x34)+(d7-)-(d7+)=2950*1.1P4:盡量避免A2到B4:x24+(d8-)-(d8+)=0B1和B3保持平衡

7、a：eox11+x21+x31-4/9(x13+x23+x33)+(d9-)-(d9+)=0目標(biāo)函數(shù)：minz=P1(d1-) + (d1+) + (d2-)+P2(d3-) + (d4-) + (d5-) + (d6-)+P3(d7+)+P4(d8+) + (d9-) + (d9+ )四、程序源代碼#include #include #include using namespace std;/*由l表Aa上?作而入業(yè)莊法的可：得i?, e?最A(yù)?小?運(yùn)?費(fèi)口?為a2950元a,倉(cāng)?其?10%為3295元3。倫 設(shè)Lxij為a工愴廠ii運(yùn)?往用?pji的i?調(diào)IjA運(yùn)?量0?, e?則“有D：

8、 eoP1:需求”約?束?： eox11+x12+x13+x14=300 x21+x22+x23+x24=200 x31+x32+x33+x34=400供?應(yīng)I約?束?： eox14+x24+x34+(d1-)-(d1+)=250供?應(yīng)I不?少，于2100個(gè)?單困位?： eox31+(d2-)-(d2+)=100P2:不?少，于以需求“的i?80%： eox11+x21+x31+(d3-)-(d3+)=160 x12+x22+x32+(d4-)-(d4+)=80 x13+x23+x33+(d5-)-(d5+)=360 x14+x24+x34+(d6-)-(d6+)=200P3： eo總A“i運(yùn)

9、?費(fèi)口?不?超?迪10%(c11*x11+c12*x12+.-.-+c34*x34)+(d7-)-(d7+)=2950*1.1P4盡?量 0? 避A1 免 aA2到i?B4:x24+(d8-)-(d8+)=0B1和aB3保Aie持?平?衡a： eox11+x21+x31-4/9(x13+x23+x33)+(d9-)-(d9+)=0目？標(biāo)入”。函數(shù)： eomin z=P1(d1-)+(d1+)+(d2-)+P2(d3-)+(d4-)+(d5-)+(d6-)+P3(d7+)+P4(d8+)+(d9-)+(d9+)*/#define N 20#define M 40#define K 10#defi

10、ne T 5#define CH 100void main()string int_into_char(int num);bool Is_number(char ch);void analyse(double *p_val1,double *p_val2,string *str_p1,string *str_p2,string *str_x1,string*str_x2,string *str_tjnt x_numjnt d_numjnt p,double (*a)Mdouble *var_val,double (*test)M);void compute(double *p_val1doub

11、le *p_val2,string *str_p1,string *str_p2,string *str_x1,string*str_x2,string *str_tjnt x_numjnt d_numjnt p,double (*a)Mdouble (*test)M);double sum_money=0,xN;for(int i=0;iN;i+) xi=0;double testKM;/*char ch1NCH,ch_funCH;memset(test,0,sizeof(test);memset(x,0,sizeof(x);double aNM,p_val1M,p_val2M,var_va

12、lN;string str_p1M,str_p2N,str_x1M,str_x2N,str_tT;int x_num,d_num,p;*/double initial_valN=5,2,6,7,3,5,4,6,4,5,2,3;double aNM=0,0,0,1,0,0,0,1,0,0,0,1,-1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,250,0,0,0,0,0,0,0,0,1,0,0,0,0,-1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,100,1,0,0,0,1,0,0,0,1,0,0,0,0,0,-1,0,0,0,0,0,0,0,0

13、,1,0,0,0,0,0,0,160,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,80,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,1,0,0,0,0,360,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,1,0,0,0,200,5,2,6,7,3,5,4,6,4,5,2,3,0,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,1,0,0,3245,0,0,0,0,0,0,0,1,0,0,0

14、,0,0,0,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,1,0,0,1,0,0.4444,0,1,0,0.4444,0,1,0,0.4444,0,0,0,0,0,0,0,0,0,-1,0,0,0,0,0,0,0,0,1,0,;double p_val1M=0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,1,1,1,1,1,1,1,1,0,0,1;double p_val2N=1,1,1,1,1,1,0,0,1;int x_num=12,d_num=9,p=4;stringstr_p1M=0,0r0,”0,”0,”0,”0,0,0,”0,”0,”0,”

15、p1,”0,”0,”0,0,0,”P(pán)3,P4,”P(pán)4,”P(pán)1,”P(pán)1”,”P(pán)2”,”P(pán)2”,”P(pán)2”,”P(pán)2”,”0”,”0”,”P(pán)4”;string str_p2N=P1 ,”P(pán)1”,”P(pán)2”,”P(pán)2” ,P2 ,P2 ,0 ,”0”,”P(pán)4”;stringstr_x1M=x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11,x12,d1+,d2+,d3+,d4+,d5+,d6+,d7+,d8+,d9+,d1-,d2-,d3-,d4-,d5-,d6-,d7-,d8-,d9-;string str_x2N=d1-,d2-,d3-,d4-,d5-,d6-,d7-,d8-,d9-;

16、string str_tT=P1 ,P2 ,P3 ,P4,;double var_valN=250,100,160,80,360,200,3245,0,0;/*cout請(qǐng)?輸。?A?x的i?個(gè)?數(shù)0y, ?d的i?個(gè)?數(shù)0y和優(yōu)?先級(jí)?p的i?個(gè)?數(shù)y(p、 id20,x+2*d40) eox_numd_nump;for(int i=1;i=p;i+)if(i=9)char temp_ch=(char)(i+48);string temp_str=p;str_ti-1=temp_str+temp_ch;elsestr_ti-1=p+int_into_char(i);for(int i=1;i=

17、x_num+2*d_num;i+)if(i=x_num)if(i=9)char temp_ch=(char)(i+48);string temp_str=x;str_x1i-1=temp_str+temp_ch;elsestr_x1i-1=x+int_into_char(i);else if(i=x_num+d_num)if(i=9)char temp_ch=(char)(i+48-x_num);string temp_str=d;str_x1i-1=temp_str+temp_ch+;elsestr_x1i-1=d+int_into_char(i-x_num)+;elseif(i=9)cha

18、r temp_ch=(char)(i+48-x_num-d_num);string temp_str=d;str_x1i-1=temp_str+temp_ch+-;elsestr_x1i-1=d+int_into_char(i-x_num-d_num)+-;for(int i=0;id_num;i+) str_x2i=str_x1i+x_num+d_num;cout請(qǐng)?輸。?入“?各A加?程-：卸eg. p1(d1-)+p2(d2-)+p2(d2+)+3p3(d3-)+5p3(d4-)endl;cout5(x1)+4(x2)+(d1-)-(d1+)=20endl;cout目？標(biāo)妃函數(shù)：/方0?

19、程.-(請(qǐng)?按a;a照?格?式?輸?入?! e?I e? ! e?)： eoch_fun;int temp_index=0;string str_p=,str_d=;double s_num=0;for(int i=0;iM;i+)p_val1i=0;str_p1i=0;for(int i=0;istrlen(ch_fun);i+)if(Is_number(ch_funi)int temp_seat=0,flag_seat=0;doif(ch_funi=.)i+;flag_seat+;elses_num=s_num*10+(int)(ch_funi)-48;i+;if(flag_seat!=0

20、) temp_seat+;while(Is_number(ch_funi);for(int i=0;itemp_seat;i+) s_num=s_num/10;flag_seat=0;if(ch_funi=(|ch_funi=) i+;if(ch_funi=p)str_p=ch_funi;i+;while(Is_number(ch_funi)str_p=str_p+ch_funi;i+;if(ch_funi=(|ch_funi=) i+;if(ch_funi=d)str_d=d;i+；dostr_d=str_d+ch_funi;i+；while(Is_number(ch_funi);str_d

21、=str_d+ch_funi;for(int j=x_num;jx_num+2*d_num;j+)if(str_d=str_x1j)if(s_num=0) p_val1j=1;else p_val1j=s_num;str_p1j=str_p;s_num=0;str_p=”;str_d=;break;for(int i=0;iN;i+)for(int j=0;jM;j+)if(j=i+x_num) aij=-1;else if(j=i+x_num+d_num) aij=1;else aij=0;cout請(qǐng)?輸?入“?加?程.-組A： eoendl;for(int i=0;ich1i;int fl

22、ag=0;double a_num=0;string str_x=”;for(int j=0;jstrlen(ch1i);j+)if(ch1ij=+,) continue;if(ch1ij=-)flag+;j+;if(Is_number(ch1ij)int temp_seat=0,flag_seat=0;doif(ch1ij=.) j+;flag_seat+;elseif(flag=0) a_num=a_num*10+(int)(ch1ij)-48;else a_num=a_num*10-(int)(ch1ij)+48;j+;flag=0;if(flag_seat!=0) temp_seat+

23、;while(Is_number(ch1ij);for(int i=0;itemp_seat;i+) a_num=a_num/10;flag_seat=0;if(ch1ij=(|ch1ij=) j+;if(ch1ij=x)str_x=x;j+;while(Is_number(ch1ij)str_x=str_x+ch1ij;j+;if(ch1ij=d) j=j+3;flag=0;if(ch1ij=(|ch1ij=) j+;if(ch1ij=)j+;if(ch1ij=-)int temp_seat=0,flag_seat=0;j+;while(Is_number(ch1ij)if(ch1ij=.)

24、 j+;flag_seat+; elseaix_num+2*d_num=aix_num+2*d_num*10-(int)(ch1ij)+48;j+；if(flag_seat!=0) temp_seat+;for(int i=0;itemp_seat;i+) aix_num+2*d_num=aix_num+2*d_num/10; flag_seat=0;continue;elseint temp_seat=0,flag_seat=0;while(Is_number(ch1ij)if(ch1ij=.) j+;flag_seat+; else aix_num+2*d_num=aix_num+2*d_

25、num*10+(int)(ch1ij)-48;j+;if(flag_seat!=0) temp_seat+;for(int i=0;itemp_seat;i+) aix_num+2*d_num=aix_num+2*d_num/10; flag_seat=0;continue;for(int k=0;kx_num+2*d_num;k+)if(str_x=str_x1k)if(a_num=0&flag=0) aik=1;else if(flag!=0&a_num=0) aik=-1;else aik=a_num;a_num=0;flag=0;str_x=”;break;for(int i=0;id

26、_num;i+)str_p2i=str_p1i+x_num+d_num;p_val2i=p_val1i+x_num+d_num;*/compute(p_val1,p_val2, str_p1, str_p2, str_x1, str_x2, str_t, x_num,d_num,p,a,test);for(int i=0;id_num;i+)for(int j=0;j0)xj=aix_num+2*d_num;else xj=0;cout最人?優(yōu)?a為a： eoendl;for(int i=0;ix_num;i+)coutsetw(5)xi+1c=xi;coutendl;for(int i=0;

27、iN;i+) sum_money+=xi*initial_vali;cout最人?優(yōu)?運(yùn)?費(fèi)口?為 asum_money元a。 jeendl; cout是。?否?進(jìn)?行D靈礦敏?度分?析?？ e?(y & n)whether_anayse;if(whether_analyse=y,)cout下?面?進(jìn)?行。靈西敏?度.分H?析？*endl;analyse(p_val1,p_val2,str_p1,str_p2,str_x1,str_x2,str_t,x_num,d_num,p,a,var_val,test);system(pause);/單g屯2?形?他static int times=1;v

28、oid compute(double *p_val1double *p_val2,string *str_p1,string *str_p2,string *str_x1,string *str_x2,string *str_tjnt x_numjnt d_numjnt p,double (*a)Mdouble (*test)M) bool is_less_than_0(int *x1_indexint x_num,int d_num,double (*test)Mjnt p);void matrixQouble(*a)Mjnt x1_index,string *str_x1,string *

29、str_x2,string *str_p1,string *str_p2,double *p_val1double *p_val2jnt d_numjnt x_num);void testnumber(string *str_p1,string *str_p 團(tuán)ouble (*a)Mdouble *p_val1double *p_val2jnt x_num,int d_numjnt p,string *str_tdouble (*test)M);void display(double(*a)M,string *str_x1,string *str_x2,string *str_p1,strin

30、g *str_p2,string *str_t,double *p_val1,double (*test)Mdouble *p_val2jnt d_numjnt x_numjnt p);int x1_index=0;testnumber(str_p1,str_p2,a,p_val1,p_val2,x_num,d_num, p,str_t,test);display(a,str_x1,str_x2,str_p1,str_p2,str_t,p_val1,test,p_val2,d_num,x_num, p);while(is_less_than_0(&x1_index,x_num,d_num,te

31、st,p)matrix(a,x1_index,str_x1,str_x2,str_p1,str_p2,p_val1,p_val2,d_num, x_num);memset(test,0,sizeof(test);for(int i=0;ip;i+)for(int j=0;jx_num+2*d_num;j+) testij=0;testnumber(str_p1,str_p2,a,p_val1,p_val2,x_num,d_num, p,str_t,test); display(a,str_x1,str_x2,str_p1,str_p2,str_t,p_val1,test,p_val2,d_nu

32、m,x_num, p);/輸?出?單tQ純2?形?表Aavoid display(double(*a)M,string *str_x1,string *str_x2,string *str_p1,string *str_p2,string *str_double *p_val1,double (*test)Mdouble *p_val2,int d_numjnt x_numjnt p) cout*NO.times次2?*endl;for(int i=0;i1) coutsetw(13)p_val1istr_p1i;else coutsetw(15)1) coutsetw(3)p_val1ist

33、r_p1i;else coutsetw(5)str_p1i;coutendl;coutsetw(5)CBsetw(5)XB”;for(int i=0;ix_num+2*d_num;i+) coutsetw(5)str_x1i;coutsetw(5)bendl;for(int j=0;jd_num;j+)if(p_val2j=1) coutsetw(5)str_p2j;else coutp_val2jsetw(4)str_p2j;coutsetw(5)str_x2j;for(int i=0;ix_num+2*d_num+1;i+) coutsetw(5)aji; coutendl;for(int

34、 i=0;ip;i+)coutsetw(10)str_ti;for(int j=0;jx_num+2*d_num;j+) coutsetw(5)testij;coutendl;times+;/計(jì)?算?檢-驗(yàn)數(shù)yvoid testnumber(string *str_p1,string *str_p免ouble (*a)Mdouble *p_val1double *p_val2jnt x_numjnt d_num,int p,string *str_t,double (*test)M)int falg1,flag2;for(int k=0;kp;k+)falg1=0;flag2=0;for(in

35、t i=0;id_num;i+)if(str_p2i=str_tk)falg1+;if(flag2=0)for(int j=0;jx_num+2*d_num;j+)if(str_p1j=str_tk) testkj=p_val1j-p_val2i*aij;else testkj-=p_val2i*aij;elsefor(int j=0;jx_num+2*d_num;j+) testkj-=p_val2i*aij;flag2+;if(falg1=0)for(int i=0;i0) min=atemp2*d_num+x_num/atempx1_indexbreak; else temp+;if(t

36、empd_num) break;for(int i=temp+1;i0&ai2*d_num+x_num/aix1_indexmin) temp=i;str_x2temp=str_x1x1_index;str_p2temp=str_p1x1_index;p_val2temp=p_val1x1_index;for(int i=0;id_num;i+)iRaix1_index=0)continue;flag=0;for(int j=0;j=2*d_num+x_num;j+)if(i!=temp)if(flag=0) temp_val=aix1_index;aij-=atempj*temp_val/a

37、tempx1_index;flag+;else break;double temp_a=atempx1_index;for(int j=0;j=2*d_num+x_num;j+)atempj=atempj/temp_a;/判D斷?檢-驗(yàn)I數(shù)y是?否口?都?小?于20bool is_less_than_0(int *x1_indexint x_num,int d_num,double (*test)M jnt p)int flag1=0,flag2=0;double min_val=0;for(int i=0;ip;i+)for(int j=0;jx_num+2*d_num;j+)if(test

38、ijtestij)for(int k=0;k0) flag2+;if(flag2=0)min_val=testij;*x1_index=j;flag1+;else flag2=0;if(flag1!=0)return true;return false;void analysedouble *p_val1,double *p_val2,string *str_p1,string *str_p2,string *str_x1,string *str_x2,string*str_t,int x_numjnt d_numjnt p,double (*a)M,double *var_valdouble

39、 (*test)M)string var_str;char go_on=y;while(go_on=,y,)cout請(qǐng)?輸。?入?你?想?加?析?的1?變A?量?jī)?e;b,c) e?： eovar_str;if(var_str=bllvar_str=B)double BNN,min=0,max=0,temp=0;double sum=0;int var_num=0;memset(B,0,sizeof(B);for(int i=0;id_num;i+)for(int j=0;jd_num;j+) Bij=aij+x_num+d_num;cout您想?分?析?第i2幾?個(gè)?變A?量?jī)?？倉(cāng)？”(0var_numd_num)var_num;for(int i=0;id_num;i+)for(int j=0;j=max) max=temp;elseif(min=0&tempmin) min=temp;else if(min0&tempmin) min=temp;coutbvar_num的 1?范口?圍ji是?： e