数据结构课程设计报告1.doc

数据结构课程设计报告一、 单词接龙程序：#include#include#include#define m 20#define n 100int main(void) char strmn; char jielong2000; int n,i; char kaitou; void fjielong(char pstrn,char *pjielong,int n_x,char kaitou_x); printf(input the number of the strings:); scanf(%d,&n); printf(input the strings:); for(i=0;in;i+) scanf(%s,stri); getchar(); printf(kaitou?); scanf(%c,&kaitou); fjielong(str,jielong,n,kaitou); printf(the final strings is:%s,jielong); system(pause); return 0;void fjielong(char pstrn,char *pjielong,int n_x,char kaitou_x) int i; for(i=0;in_x;i+) if(pstri0=kaitou_x) strcpy(pjielong,pstri); pjielong+=strlen(pstri+1); break; for(i=0;in_x;i+) if(pstri0!=kaitou_x) if(*(pjielong-1)= pstri0) pjielong-; strcpy(pjielong,pstri); pjielong+=strlen(pstri+1); return 0;运行结果：二、方格问题程序：main()int i,n,g,j;int f1101=0,f2101=0;int f3101;scanf(%d,&n);f1100=1;f2100=3;for(i=3;i=1;j-)g=g+2*f1j+f2j;f3j=g%10;g=g/10;for(j=1;j=100;j+)f1j=f2j;f2j=f3j;j=1;while(f3j=0)j+;for(i=j;i=100;i+)printf(%d,f3i);printf(n);运行结果二、 fbz串程序：main()int i,j,k,n,temp;int s10002;char q2000=0;temp=1;scanf(%d,&k);if(k=1) n=1;else for(i=2;i=k;i+) n=temp*2;temp=n;for(i=0;i=n-1;i+)for(j=0;j=1;j+)scanf(%d,&sij);for(i=0;i=n-1;i+)for(j=0;j=1;j+)if(sij=0) qn*2+i*2+j=z; else qn*2+i*2+j=b;for(i=0;i=2;j=j/2)for(i=0;i=j-1;i=i+2)if(qj+i=qj+i+1)&(qj+i!=f)q(j+i)/2=qj+i; else q(j+i)/2=f;printf(%c,q1);for(i=1;i=n*2-1;i=i+)if(qi=f)printf(%c%c,qi*2,qi*2+1);getch();运行结果：四、 分油程序：main () int q1004=0,0,0,0,10,0,0,0; int j,front,rear,x10,x7,x3,y10,y7,y3,p,p1,i; front=1;rear=1;p=1; while(p) x10=qfront0; x7=qfront1; x3=qfront2; front+; if(x10=5)&(x7=5) p=0; else if(x77) y10=x10+x7-7;y7=7;y3=x3; p1=1; for (i=1;i=rear;i+) if (y10=qi0)&(y7=qi1) p1=0; if (p1=1) rear+; qrear0=y10; qrear1=y7; qrear2=y3; qrear3=front-1; if (x33) y10=x10+x3-3; y7=x7; y3=3; p1=1; for (i=1;i0) y10=x10+x7; y7=0; y3=x3; p1=1; for (i=1;i0)&(x33) y7=x7+x3-3;y3=3; else y7=0;y3=x7+x3; p1=1; for (i=1;i0) y10=x10+x3; y7=x7; y3=0; p1=1; for (i=1;i=rear;i+) if (y10=qi0)&(y7=qi1) p1=0; if (p1=1) rear+; qrear0=y10; qrear1=y7; qrear2=y3; qrear3=front-1; if (x70) y10=x10; if (x7+x37)y7=7;y3=x7+x3-7;elsey7=x7+x3;y3=0; p1=1; for (i=1;i=rear;i+) if (y10=qi0)&(y7=qi1) p1=0; if (p1=1) rear+; qrear0=y10; qrear1=y7; qrear2=y3; qrear3=front-1; front-; for(j=1;j=front;j+) printf (%d %d %d n,qj0,qj1,qj2); getch();运行结果：五、 关键点和桥程序：#include stdio.hmain()int g1010=0,0,0,0,0,0,0,0,0,0, 0,0,1,1,1,0,0,0,0,0, 0,1,0,1,0,0,1,0,0,0, 0,1,1,0,1,1,0,0,0,0, 0,1,0,1,0,1,0,0,0,0, 0,0,0,1,1,0,1,0,0,0, 0,0,1,0,0,1,0,1,0,0, 0,0,0,0,0,0,1,0,1,0, 0,0,0,0,0,0,0,1,0,1, 0,0,0,0,0,0,0,0,1,0;int g01010=0;int i,j,x,front,rear,s,n;int a,b;int flag10=0;int q10;n=9;printf(关键点为:);for(i=5;in;i+)/关键点检索g01010=g1010;g0ii+1=0;g0i+1i=0;flagi=1;q1=i;front=1;rear=1;while(front=rear)x=qfront;front+;for(j=1;j=n;j+)if(flagj=0)&(gxj=1)rear+;qrear=j;flagj=1;if(rear=n-1)printf( %d,i);/广度优先循环搜索判断关键点，不符合条件记录输出。printf(n);printf(桥为：);for(a=6;an;a+)/桥检索g01010=g1010;g0aa+1=0;g0a+1a=0;flaga=1;q1=a;front=1;rear=1;while(front=rear)x=qfront;front+;for(b=1;b=n;b+)if(flagb=0)&(gxb=1)rear+;qrear=b;flagb=1;if(rearn-1)printf( %d%d,a,a+1);getch();运行结果：关键点为： 6 7 8桥为： 67 78 89六、 矩阵程序：main() int i,j,n,k,n1,s,s0,s1,x,y,p; int g2020; int b20; scanf(%d,&n); k=0; for(i=0;in;i+) k+; for(j=i;ji;j-) gn-i-1j=k; gjn-i-1=k; for(i=0;in;i+) for(j=0;jn;j+) printf(%d ,gij); printf(n); s0=0; n1=n+n-2; for(i=0;in;i+) bi=0; for(i=n;in+n-1;i+) bi=1; while(b0=0) s1=1;x=1;y=1; for(i=1;i=n1;i+) if(bi=0) x+; else y+; s1=s1+gxy; p=1; j=2; while(p)&(js1) if(s1%j=0) p=0; else j+; if(p) s0+; if ( bs1) s0+; j=n1; while(bj=0) j-; s1=0; while(bj=1) s1+; bj=0;j-; bj=1; s1-; k=n; for(i=1;i=s1;i+) bk=1; k-; printf(%d,s0); getch();运行结果：七、骑士问题程序：main() int i,k,m,n,top,x,y,w,q,x0,y0; int dx9=0,1,2,2,1,-1,-2,-2,-1; int dy9=0,-2,-1,1,2,2,1,-1,-2; int stack1000; int g100100=0; scanf(%d,%d,%d,&n,&x0,&y0); while(n%2=1)&(x0+y0)%2=1) printf(input error); scanf(%d,%d,%d,&n,&x0,&y0);gx0y0=1;x=x0;y=y0;k=0;top=0; while(top9) k=stacktop; top-; gxy=0; x=x-dxk; y=y-dyk; else w=x+dxk; q=y+dyk; if(w=1)&(w=1)&(q=n)&(gwq=0) x=w; y=q; top+; gxy=1; stacktop=k; k=0; x=x0; y=y0; printf(%d %d,x,y); for(i=1;i=top;i+) x=x+dxstacki; y=y+dystacki; printf(-%d %d,x,y); getch();运行结果：八、士兵排队程序：main()int x,y,i,i1,j,k,s,s1,n,p,j1,j2; int q100; int g77=0,0,0,0,0,0,0, 0,0,0,0,0,0,1, 0,0,0,0,1,1,0, 0,0,0,0,0,0,0, 0,1,0,0,0,1,0, 0,1,0,1,0,0,1, 0,0,0,1,0,0,0; n=6; k=9; p=1;j1=1; while(p)&(j1=n) s=0; for(i=1;i=n;i+) s1=0; for (j=1;j1) printf(error2); p=0; else qj1=j2; j1+; for(i=1;i=n;i+) gij2=1; for(i=1;i=n;i+) gj2i=0; printf(n); if(p) for(i=1;i=n;i+) printf(%d ,qi); getch(); 运行结果：九、huffman树编码程序：#include stdio.h#include stdlib.hint main() int i,i0,i1,j,j1,j0,n,n1,n2,cmin1,cmin,s,k,p; char str30=wabcddeacefgfg; int tree5213=0; int h27=0; int d27; int code2721; str0=13; scanf(%d,&p); for(i=1;i=str0;i+) hstri-96+; n=0; for(i=1;i0) n+; treen1=hi; dn=i; if(n0) n1=n+(p-1)-n2; s=1+(n1-p)/(p-1); n2=n1; for(i=1;i=s;i+) n2+;cmin1=0; for(j=1;j=p;j+) cmin=1000; for(k=1;kn2;k+) if(treek0=0)&(treek1cmin) j1=k;cmin=treek1; cmin1=cmin1+cmin; treen22+p=j1; treej12=n2; treej10=1; treen21=cmin1; for(i=0;i=26;i+) for(k=0;k=20;k+) codeik=-1; for(i=1;i=n;i+) i0=i;i1=treei2;k=21; while(i1!=0) k-; for(j1=3;j13+p;j1+) if(treei1j1=i0) codeik=j1-3; i0=i1;i1=treei12; for(i=1;i=n;i+) printf(%d,di+96); k=1; while(codeik=-1) k+; for(j1=k;j1=20;j1+) printf(%d,codeij1); printf(n); getchar();system(pause); return 0;运行结果：十、最长公共子串程序：main()int i,j,cmax,j1; char a100=wabcdefghijk; char b100=wabcyefghi; char g100100=0; a0=11; b0=9; cmax=0; for(i=1;i=a0;i+) for(j=1;jcmax) j1=j; cmax=gij; else gij=0; printf(%dn,cmax); for(i=j1-cmax+1;i8)/八个方向k=stacktop;top-;gxy=0;x=x-dxk;y=y-dyk;elsew=x+dxk;q=y+dyk;if(w=0)&(q=0)&(gwq=0)/在范围内，路径可走，没做过x=w;y=q;gxy=1;top+;/flagwq=1;stacktop=k;k=0; if (x=0) & (y=n) p=0; /stacktop=k;/k=0; /printf(x=%d&y=%d)n,x,y); x=0;y=0;count=0;printf(n); for(i=0;i=top;i+)x=x+dxstacki;y=y+dystacki;count+;/printf(,x,y);/bi=;/ai=x;/bi=y;printf(,x,y); printf(最短路径是%d,count); getch(); 运行结果：十二、挖宝藏程序：#include stdio.h#define m 10000main()int i,j,k,top,cmax,cen,tk,n,t1,top1;int w20=9,36,41,28,35,53,17,21,32,36;int t1010=0,0,0,0,0,0,0,0,0,0, 0,0,13,m,11,m,9,12,m,m,/1 0,13,0,10,12,m,m,m,m,m,/2 0,m,10,0,m,13,m,m,m,m,/3 0,11,12,m,0,9,m,m,8,m,/4 0,m,m,13,9,0,m,m,m,14,/5 0,9,m,m,m,m,0,17,m,m,/6 0,12,m,m,m,m,17,0,10,m,/7 0,m,m,m,8,m,m,10,0,11, 0,m,m,m,m,14,m,m,11,0; int flag10=0;int stack100;k=40;n=9;cmax=w1;flag1=0;cen=w1;top=0;stack0=1;k=0;t1=0;while(top=0)k+;if(kn)k=stacktop;top-;cen=cen-wk;flagk=0;t1=t1-tstacktopk;else if(flagk=0)&(t1+tstacktopk)flagk=1;cen=cen+wk