递归方法实现顿回路问题

//递归方法实现回路问#include<string.h>#include<iostream.h>boolc[6][6];intx[6];intvoidoutput(int{for(inti=1;i<=n;i++)cout<<x[i]<<'}voidswap(int&x,int{intt=x;x=y;y=}voidhamilton(intk{if(k>n&&c[x[k-1]][x[1]]==1)//一个解{for(int{if(c[x[k-1]][x[k]]==1)hamilton(k+1);}}}int{intfor(i=1;i<6;++i)x[i]=i;return0;intn=memset(c,0,c[1][2]=c[2][1]=c[1][4]=c[4][1]=c[2][4]=c[4][2]=c[2][3]=c[3][2]=c[3][5]=c[5][3]=c[4][5]=c[5][4]=c[2][5]=c[5][2]=c[3][4]=c[4][3]=for(inti=1;i<=n;{x[i]=}cout<<endl;return}//回路问题回溯法，解空间树按子集树构造#include<string.h>#include<iostream.h>voidhamilton(intn,intx[],bool{inti,bool*visited=newbool[n+1];for(i=1;i<=n;i++){x[i]=visited[i]=};//while(k{while(x[k]<={if(visited[x[k]]==0&&c[x[k1]][x[k]]==1)//未到，有边}

if(x[k]n&&k==n&&c[x[k]][x[1]]==1)//{cout<<x[k]<<''; break;}elseifx[k]<=n&&k<n)//{}else{}}delete}int{intn5五个城市！boolc[6][6];memset(c,0,c[1][2]=c[2][1]=c[1][4]=c[4][1]=c[2][4]=c[4][2]=c[2][3]=c[3][2]=c[3][5]=c[5][3]=c[4][5]=c[5][4]=c[2][5]=c[5][2]=c[3][4]=c[4][3]=true;intx[6];hamilton(5,x,c);cout<<endl;return0;}#include<stdio.h>#includestring.h>#includestdlib.h>#includeintmain(){#ifdefDEBUGintbase[]=intk=sizeof(base)/sizeof(int); unsignedlongintf[10][50001];intintfor(i=0;i<50001;++i)f[0][i]=for(i=0;i<10;++i)f[i][0]=1;for(i=1;i<10;{for(j=1;j<50001;{f[i][j]=f[i-}}

charwhile(gets(msg)!= &&strcmp(msg,"0"){p=intcount={count+=p=}}

#ifdefDEBUGend(FILE);return}#include<stdio.h>#includestring.h>#includestdlib.h>#includeintmain(){#ifdefDEBUGinta[11];/*intbase[]={0,1,5,10,20,50,100,200,500,1000,2000};*/intbase[]={0,2000,1000,500,200,100,50,20,10,5,1};intk=sizeof(base)/sizeof(int)-1; intx,m;intintlines=charwhile(gets(msg)!= &&strcmp(msg,"0"){p=intmaxnum=0;intcount=0;lines={x=atoi(str);a[lines]=x;count+=x;lines--;p=}intintintrow=1;sp[row]=-base[row];s[0]=count;s[1]=longlongintl=0;{sp[row]+=sp[row]=s[row];{{rowsp[row]=-}}else{}{}}

row--}#ifdefDEBUGend(FILE);return}回溯递归//nm，图的邻接矩阵//输出n个顶点的#include<iostream.h>boolc[6][6];intx[6];intm=3;//颜色数intn=5;boolok(intk)//判断对顶点k以后是否合{intfor(i=1;i<k;if((c[k][i]==1&&x[k]==x[i]))returnfalse;return}voidoutput(int{for(inti=1;i<=n;i++)cout<<x[i]<<'}voidbacktrack(int{if(t>n)output(x);for(inti=1;i<=m;i++){if(ok(t))backtrack(t+1);}}int{inti,for(i=1;i<5;for(j=1;j<5;j++)c[i][j]=c[1][2]=c[1][3]=c[2][3]=c[2][4]=c[2][5]=c[3][5]=c[4][5]=c[2][1]=c[3][1]=c[3][2]=c[4][2]=c[5][2]=c[5][3]=c[5][4]=}}cout<<endl;return0;}回溯迭代//nm，图的邻接矩阵//输出n个顶点的//第一个结点也可以有m种方#include<iostream.h>boolc[6][6];intx[6];intm=3;intn=5;boolok(intk)//判断对顶点k以后是否合{intfor(i=1;i<k;if((c[k][i]==1&&x[k]==x[i]))returnfalse;return}voidm_coloring(intn,int{inti,for(i=1;i<=n;i++)x[i]=0;kwhile(k{while(x[k]<={if(ok(k)==1)break;elsex[k]++;if(x[k]m&&k==n){cout<<x[i]<<""; }elseif(x[k]<=m&&k<n){}}}int{inti,for(i=1;i<5;for(j=1;j<5;j++)c[i][j]=c[1][2