北邮 算法设计与分析_第三章实验报告

算法设计与分析第三章程序作业源程序代码1.最长公共子序列#include<stdio.h>#include<stdlib.h>#define N 1000int e[N][N],f[N][N];void LCSLength(int m,int n,char *x,char *y,int c[][N],int b[][N])//构造数组b[i][j]{int i,j;for (i=1; i<=m;i++)c[i][0]=0; //初始化, Y[j]为空时for (i=1;i<=n;i++)c[0][i]=0; //初始化，X[i]为空时for (i=1;i<=m;i++) //两重循环，自下而上，// 计算子问题{X(i), Y(j)}for (j=1;j<=n;j++){if(x[i]==y[j]){ //情况1c[i][j]=c[i-1][j-1]+1;b[i][j]=1;}else if(c[i-1][j]>=c[i][j-1]){ //情况2c[i][j]=c[i-1][j];b[i][j]=2;}else //情况3{c[i][j]=c[i][j-1];b[i][j]=3;}}}void LCS(int i,int j,char *x,int b[][N]){if (i==0||j==0)return;if (b[i][j]==1){LCS(i-1,j-1,x,b);printf("%c",x[i]);/*第1种情况下，X(i)和Y(j)的最长公共子序列由X(i-1)和Y(j-1)的解LCS(i-1, j-1, x, b)，加上位于最后的X[i]组成*/}else if(b[i][j]==2)LCS(i-1,j,x,b);elseLCS(i,j-1,x,b); //其它2种情况下，原问题解等于子问题解}main(){char A[N],B[N],C[N],D[N];int a=1,b=1,c=1,d=1;char ch;int i;FILE * fp;if((fp=fopen("最长公共子序列输入数据.txt","r"))==NULL)//打开文件失败printf("Can't open file ");else{//读取文件内容ch=fgetc(fp);ch=fgetc(fp);ch=fgetc(fp);ch=fgetc(fp);ch=fgetc(fp);while(ch!='B'&&ch!=EOF) {A[a]=ch;a++;ch=fgetc(fp);while(ch=='\n')ch=fgetc(fp);}a--;ch=fgetc(fp);ch=fgetc(fp);ch=fgetc(fp);ch=fgetc(fp);while(ch!='C'&&ch!=EOF) {B[b]=ch;b++;ch=fgetc(fp);while(ch=='\n')ch=fgetc(fp);}b--;ch=fgetc(fp);ch=fgetc(fp);ch=fgetc(fp);ch=fgetc(fp);while(ch!='D'&&ch!=EOF) {C[c]=ch;c++;ch=fgetc(fp);while(ch=='\n')ch=fgetc(fp);}c--;ch=fgetc(fp);ch=fgetc(fp);ch=fgetc(fp);ch=fgetc(fp);while(ch!=EOF){D[d]=ch;d++;ch=fgetc(fp);while(ch=='\n')ch=fgetc(fp);}d--;}fclose(fp);printf("文件读取完成\n");/* for(i=0;i<=a;i++)printf("%c",A[i]);printf("\n");for(i=0;i<=b;i++)printf("%c",B[i]);printf("\n");for(i=0;i<=c;i++)printf("%c",C[i]);printf("\n");for(i=0;i<=d;i++)printf("%c",D[i]);printf("\n"); */printf("\n\n\nA和B的最长公共子序列为：\n");LCSLength(a,b,A,B,e,f);LCS(a,b,A,f);printf("\n\n\nC和D的最长公共子序列为：\n");LCSLength(c,d,C,D,e,f);LCS(c,d,C,f);printf("\n\n\nA和D的最长公共子序列为：\n");LCSLength(a,d,A,D,e,f);LCS(a,d,A,f);system("pause");return 0;}2.最大子段和#include<stdio.h>#include<stdlib.h>#define N 400void MaxSum(int n,int *a){int sum=0,b=0;int i,x1=0,y1=0,x=0,y=0;for(i=1;i<=n;i++){if (b>0){b+=a[i];y1=i;}else{b=a[i];x1=i;}if(b>sum){sum=b;x=x1;y=y1;}}printf("\n该序列的最大子段和的位置为从第%d个到第%d个",x+1,y+1);printf("\n该序列的和最大子段为：\n");for(i=x;i<=y;i++)printf("%d ",a[i]);printf("\n它们的和为%d \n",sum);}main(){int A[N],B[N];int a=0,b=0;int i;FILE *fp1,*fp2;if((fp1=fopen("最大子段和输入数据-序列1.txt","r"))==NULL)//打开文件失败printf("Can't open file1 ");else //读取文件内容{for(i=0;!feof(fp1);i++)fscanf(fp1,"%d\n",&A[i]);a=i-1;}if((fp2=fopen("最大子段和输入数据-序列2.txt","r"))==NULL)//打开文件失败printf("Can't open file2 ");else{for(i=0;!feof(fp2);i++)fscanf(fp2,"%d\n",&B[i]);b=i-1;}fclose(fp1);fclose(fp1);printf("文件读取成功！！！\n\n\n");printf("对于最大子段和输入数据-序列1 ");MaxSum(a,A);//计算最大子段和printf("\n\n对于最大子段和输入数据-序列2 ");MaxSum(b,B);// for(i=0;i<=b;i++)// printf("%d ",B[i]);system("pause");return 0;}3.凸多边形最优三角剖分#include<stdio.h>#include<stdlib.h>#include<math.h>#define N 30#define EARTH_RADIUS 6378.137struct Base{//基站结构体int ENODEBID;//基站标识float LONGITUDE;//基站经度float LATITUDE;//基站纬度int NUM;//基站k-dist距离};void readfile(struct Base * base,int n) //读取文件函数{FILE * fp;int i=0;if(n==21)fp=fopen("21个基站凸多边形数据.csv","rb");elsefp=fopen("29个基站凸多边形数据.csv","rb");if(fp==NULL)//打开文件失败printf("Can't open file ");else{while(EOF!=fscanf(fp,"%d,%f,%f,%d",&base[i].ENODEBID,&base[i].LO NGITUDE,&base[i].LATITUDE,&base[i].NUM)){//将数据读入结构体数组i++;//printf("%d\n",i);}// printf("文件读入成功！！！\n\n\n");}fclose(fp);}float rad(float LatOrLon){return LatOrLon * 3.14/180.0;}float distance(struct Base * base,int a,int b)//求两个基站间的距离{float s;float radLat1,radLat2,radlng1,radlng2;radLat1 = rad(base[a].LATITUDE);radLat2 = rad(base[b].LATITUDE);radlng1 = rad(base[a].LONGITUDE);radlng2 = rad(base[b].LONGITUDE);//利用正弦余弦公式求距离if(radLat1==radLat2&&radlng1==radlng2)s=0;else{s=acos(cos(radLat1)*cos(radLat2)*cos(radlng1-radlng2)+sin(radLat1)*s in(radLat2));s = s * EARTH_RADIUS;s=round(s*1000);}return s;}float weight(struct Base * base,int a,int b,int c){float s,s1,s2,s3;s1=distance(base,a,b);s2=distance(base,b,c);s3=distance(base,a,c);//s=distance(base,a,b)+distance(base,b,c)+distance(base,a,c);s=s1+s2+s3;return s;}float MinWeightTriangulation(struct Base * base,int n,float t[][N],int s[][N]){int i,j,r,k;float u=0;for(i=1;i<=n;i++)t[i][i]=0;for(r=2;r<=n;r++) //r为当前计算的链长（子问题规模）for(i=1;i<=n-r+1;i++)//n-r+1为最后一个r链的前边界{j=i+r-1;//计算前边界为r，链长为r的链的后边界t[i][j]=t[i+1][j]+weight(base,i-1,i,j);//将链ij划分为A(i) * ( A[i+1:j] )这里实际上就是k=is[i][j]=i;for (k=i+1;k<i+r-1;k++) //将链ij划分为( A[i:k] )* (A[k+1:j]){u=t[i][k]+t[k+1][j]+weight(base,i-1,k,j);if(u<t[i][j]){t[i][j]=u;s[i][j]=k;}}}return t[1][n];}void Traceback(int i,int j,int s[][N]){if(i==j)return;Traceback(i,s[i][j],s);Traceback(s[i][j]+1,j,s);printf("三角剖分顶点：V%d , V%d , V%d\n",i-1,j,s[i][j]); }main(){int i;float t[N][N];int s[N][N];struct Base base21[N];//结构体（N宏定义）struct Base base29[N];//结构体（N宏定义）readfile(base21,21);//将数据从文件中读到结构体中readfile(base29,29);//将数据从文件中读到结构体中printf("文件读入成功！！！\n\n\n");// for(i=0;i<=28;i++)//printf("%6d %.3f %.5f %d\n",base29[i].ENODEBID,base29[i].LONGITU DE,base29[i].LATITUDE,base29[i].NUM);printf("凸21多边形的最优三角剖分值为%f",MinWeightTriangulation(base21,20,t,s));printf("最优三角剖分结构为：\n");Traceback(1,20,s); //s[i][j]记录了Vi-1和Vj构成三角形的第3个顶点的位置printf("凸29多边形的最优三角剖分值为%f",MinWeightTriangulation(base29,28,t,s));printf("最优三角剖分结构为：\n");Traceback(1,28,s); //s[i][j]记录了Vi-1和Vj构成三角形的第3个顶点的位置system("pause");return 0;}4.01背包问题#include<stdio.h>#include<stdlib.h>#define N 100struct Goods{int weight;//物品重量int value;//物品价值};int knapsack(struct Goods *goods,int bag,int number,int p[][2],int head[]){int left=0,right=0,next=1,i,j,k,m,y;p[0][0]=0;p[0][1]=0;head[number+1]=0;head[number]=1;for(i=number;i>=1;i--){k=left;for(j=left;j<=right&&p[j][0]+goods[i-1].weight<=bag;j++){ y=p[j][0]+goods[i-1].weight;m=p[j][1]+goods[i-1].value;while(k<=right&&p[k][0]<y){p[next][0]=p[k][0];p[next][1]=p[k][1];next++;k++;}if(k<=right&&p[k][0]==y){if(m<p[k][1]){m=p[k][1];}k++;}if(m>p[next-1][1]){p[next][0]=y;p[next][1]=m;next++;}while(k<=right&&p[k][1]<=p[next-1][1]){k++;}}while(k<=right){p[next][0]=p[k][0];p[next][1]=p[k][1];next++;k++;}left=right+1;right=next-1;head[i-1]=next;}return next;}void traceback(struct Goods *goods,int number,int head[],int p[][2],int x[]){int j=p[head[0]-1][0],m=p[head[0]-1][1];int i,k,a;for(i=1;i<=number;i++){x[i]=0;a=1;// for(k=3450;k<=head[i]-1&&a==1;k++)for(k=head[i+1];k<=head[i]-1&&a==1;k++){if(p[k][0]+goods[i-1].weight==j&&p[k][1]+goods[i-1].value==m){x[i]=1;j=p[k][0];m=p[k][1];a=0;}}}}main(){int bag1,bag2;//背包容量int number1,number2;//物品数目int i,n;int p[20000][2];int h[N],x[N];struct Goods goods1[N],goods2[N];FILE *fp;char ch='0';if((fp=fopen("附件4.背包问题输入数据.txt","r"))==NULL)//打开文件失败printf("Can't open file ");else //读入背包和物品信息{fseek(fp,18,SEEK_CUR);fscanf(fp,"%d",&bag1);//背包1容量fseek(fp,18,SEEK_CUR);ch='0';for(i=0;ch!='\n';i++){ch=fgetc(fp);}number1=i;ch='0';fseek(fp,14,SEEK_CUR);for(i=0;ch!='\n'&&ch!=EOF;i++){fscanf(fp,"%d",&goods1[i].value);//物品价值ch=fgetc(fp);}if(number1!=i){printf("读取数据出错，程序将退出！\n");system("pause");}fseek(fp,22,SEEK_CUR);fscanf(fp,"%d",&bag2);//背包1容量fseek(fp,18,SEEK_CUR);ch='0';for(i=0;ch!='\n';i++){ch=fgetc(fp);}number2=i;ch='0';fseek(fp,14,SEEK_CUR);for(i=0;ch!='\n'&&ch!=EOF;i++){fscanf(fp,"%d",&goods2[i].value);//物品价值ch=fgetc(fp);}if(number2!=i){printf("读取数据出错，程序将退出！\n");system("pause");}}printf("文件读取成功！！！\n");/*打印文件的信息*/printf("\n\n第一组背包数据：\n");printf("背包容量：%d 物品数量：%d\n物品重量：\n",bag1,number1);for(i=0;i<number1;i++)printf("%d ",goods1[i].weight);printf("\n物品价值：\n");for(i=0;i<number1;i++)printf("%d ",goods1[i].value);printf("\n\n第二组背包数据：\n");printf("背包容量：%d 物品数量：%d\n物品重量：\n",bag2,number2);for(i=0;i<number2;i++)printf("%d ",goods2[i].weight);printf("\n物品价值：\n");for(i=0;i<number2;i++)printf("%d ",goods2[i].value);n=knapsack(goods1,bag1,number1,p,h);traceback(goods1,number1,h,p,x); //确定放入的物品//打印结果printf("\n\n第一组：\n背包重量为：%d 背包价值为：%d\n",p[n-1][0],p[n-1][1]);printf("放入的物品为：\n");for(i=1;i<number1;i++)if(x[i]==1)printf("物品%-2d 重量%-2d 价值%d\n",i,goods1[i-1].weight,goods1[i-1].value);//第二组数据n=knapsack(goods2,bag2,number2,p,h);traceback(goods2,number2,h,p,x);printf("\n\n第二组：\n背包重量为：%d 背包价值为：%d\n",p[n-1][0],p[n-1][1]);printf("放入的物品为：\n");for(i=1;i<number2;i++)if(x[i]==1)printf("物品%-2d 重量%-2d 价值%d\n",i,goods2[i-1].weight,goods2[i-1].value);system("pause");return 0;}运行结果1.最长公共子序列2.最大子段和3.凸多边形最优三角剖分4.01背包问题。