計(jì)算機(jī)算法設(shè)計(jì)與分析試題答案

1、、子集樹(shù)：void backtrack(t)if(tn) output(x)elsefor(i=0;in) output(x)elsefor(i=t;i=N1，有 F（N）=N2 有 G(N)=N3 有F(N)=c1f(N)=c1h(N)=c3h(N)類(lèi)似的有G(N)=c2f(N)=v2h(N)=c3h(N)因而O(f)+O(g)=F(N)+G(N)=c3h(N)+c3h(N)=sc3h(N)=O(h)=O(max(f,g)、t(n)=3t(n/2)+cn=33t(n/22)+cn/2+cn=32t(n/22)=3it(n/2i)=3it (n/2i)+2(3i/2i-1)cn令 n/2i=1

2、，則 i=logn所以 t(n)=O(nlog3)、加括號(hào)為(A1(A2(A3(A4A5)或者(A1(A2A3)(A4A5)、2357557979101291210122212161638A:10、b:011c:0101 d:0100e:00 f:11一般算法是 O(n3)和 O(n2)。分治法是 O(nlogn)動(dòng)態(tài)規(guī)劃算法是 O(n),動(dòng)態(tài)規(guī)劃算法如下：maxsum(a,n)/*a為 n 個(gè)整數(shù)的數(shù)組，n 為數(shù)組的大小*/sum=0,b=0,i; for(i=0;i0) b+=ai; elseb=ai; if(bsum) sum=b;Return sum;、a、五個(gè)逆序?qū)Ψ謩e是：b、當(dāng)集合

3、的順序?yàn)閚,n-1,n-2,1時(shí)有最多的逆序?qū)?，最多?n(n-1)/2 個(gè)。#include #include count(a,b,l,m,r);01或04003000175040002400010002500180003000100006000void copyback(a,a,b,b,left,left,right);right);inver(inver(a,b,left,right)mid,num=0,leftnum=0,rightnum=0; if(leftright)mid=(left+right)/2; leftnum=inver(a,b,left,mid);rightnum=i

4、nver(a,b,mid+1,right);num=count(a,b,left,mid,right);copyback(a,b,left,right);return leftnum+rightnum+num;count(a,b,l,m,r)i=l,j=m+1,k=l,q,num=0; while(i=m)&(j=r)if(aim)for(q=j;q=r;q+) bk+=aq;elsefor(q=i;q=m;q+) bk+=aq;num+=(m-i)*(r-m);return num;void copyback(i;a,b,left,right)for(i=left;i=right;i+)ai=bi;void main()a=5,1,2,4,3,i,num;n=sizeof(a)/sizeof();*b;b=(*)malloc(size