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哈密顿回路汇编语言实现(有小bug)

忠告:本代码是有bug的,有一些平行边,单点之类的情况好像没有考虑到,C++代码以及汇编代码只是给大家一个对于递归汇编程序直观的印象,并不是std,且本文的汇编实现
十分不好,是笔者年轻时写的一坨东西。关于更成熟的汇编语言实现递归,请参考笔者P2的那篇文章中的全排列以及汉诺塔部分。

C++实现
#include <iostream> #include <cstdio> #include <bits/stdc++.h> /*邻接矩阵存储图,从起点开始,直接深搜,打标记,一直搜到不可以再搜,看看是不是走了n步,并且头尾相连*/ int graph[30][30]; int sign[30]; int n,m,ans=0; void DFS(int now,int step); int main() { int u,v; scanf("%d %d",&n,&m); //不妨图的顶点序号从1开始 for(int i=0;i<m;i++){ scanf("%d %d",&u,&v); graph[u][v]=graph[v][u]=1; } sign[1]=1; DFS(1,1); printf("%d\n",ans); return 0; } void DFS(int now,int step){ if(step==n){ if(graph[1][now]){ ans=1; } return; } for(int i=1;i<=n;i++){ if(!sign[i] && i!=now && graph[now][i]){ sign[i]=1; DFS (i,step+1); sign[i]=0; } } return; }
汇编实现:(有小bug,似乎会被卡单点或者两点,但特判之后反而错得更多了)
代码已做特殊处理
.macro done li $v0,
10 syscall .end_macro ############################################################# .macro initial li $t0,0 for_1: beq $t0,$s0,for_1_end sll $t1,$t0,2 #t1=t0*4(int is 4 byte) add $t1,$t1,$s3 sw $0,0($t1) #sign[t1]=0 addi $t0,$t0,1 #t0=t0+1 j for_1 for_1_end: .end_macro ############################################################# .macro getindex(%ans,%i,%j) mult %i,$t6 #i*10 mflo $t4 #t4=i*10 add %ans,$t4,%j #t4=i*10+j sll %ans,%ans,2 #ans=ans*4 .end_macro ############################################################# .data matrix: .space 400 sign: .space 40 ######################################################### .text main: la $s2,matrix la $s3,sign li $t6,10 #tmp li $t7,1 #tmp li $s7,0 #s7 is answer li $v0,5 syscall move $s0,$v0 #s0 is n li $v0,5 syscall move $s1,$v0 #s1 is m initial li $t0,0 #t0 is i for_create: beq $t0,$s1,for_create_end #if i==m,stop add edge li $v0,5 syscall move $t1,$v0 #t1 is u li $v0,5 syscall move $t2,$v0 #t2 is v getindex($t3,$t1,$t2) #get the address in matrix add $t3,$t3,$s2 #get address in matrix,t3 is address sw $t7,0($t3) #add edge (u,v) getindex($t3,$t2,$t1) add $t3,$t3,$s2 sw $t7,0($t3) #add edge (v,u) addi $t0,$t0,1 #i++ j for_create for_create_end: sw $t7,4($s3) #sign[1]=1 li $a0,1 #a0 is now position li $a1,1 #a1 is step jal dfs #the return address is in $31 li $v0,1 li $a0,0 syscall done dfs: sw $31,0($sp) addi $sp,$sp,-4 blt $a1,$s0,dfs_else getindex($t3,$t7,$a0) add $t3,$t3,$s2 #t3 is address in matrix lw $t4,0($t3) #t4=graph[1][now] beq $t4,1,arrive_ans addi $sp,$sp,4 lw $31,0($sp) jr $31 #if we don't find the ans,return #omit something………… dfs_else: jal dfs_work addi $sp,$sp,4 lw $31,0($sp) jr $31 then: jr $31 #you have jumped to main succesfully arrive_ans: li $s7,1 li $a0,1 li $v0,1 #print answer syscall done j then dfs_work: #then ,we should do something li $t0,1 #t0 is i sw $31,0($sp) #save return address addi $sp,$sp,-4 for_dfs1: bgt $t0,$s0,for_dfs1_end bge $a1,$s0,for_dfs1_end sw $a0,0($sp) #save now addi $sp,$sp,-4 sw $a1,0($sp) #save step addi $sp,$sp,-4 #all the info was saved sucessfully sll $t1,$t0,2 #t1=t0*4 add $t1,$t1,$s3 #t1 is the address in array_sign lw $t2,0($t1) #t2 is sign[i] getindex($t3,$a0,$t0) #t3 is [now][i] lw $t3,0($t3) #t3 is graph[now][i] beq $t2,1,else #if sign[i]==1,return beq $t0,$a0,else #if i==now,return beq $t3,0,else #if graph[now][i]==0,return sw $t7,0($t1) #sign[i]=1; sw $t0,0($sp) #save tmpi in stack addi $sp,$sp,-4 sw $t1,0($sp) #save the address in array_sign addi $sp,$sp,-4 move $a0,$t0 #now=i addi $a1,$a1,1 #step++ jal dfs #next recursion addi $sp,$sp,4 lw $t1,0($sp) #fetch the address in array_sign addi $sp,$sp,4 lw $t0,0($sp) #fetch i sw $t7,0($t1) #sign[i]=0 addi $sp,$sp,4 lw $a1,0($sp) addi $sp,$sp,4 lw $a0,0($sp) #fetch all the things addi $t0,$t0,1 #i++ j for_dfs1 for_dfs1_end: addi $sp,$sp,4 lw $31,0($sp) #fetch the leeway jr $31 else: addi $t0,$t0,1 addi $sp,$sp,4 lw $a1,0($sp) addi $sp,$sp,4 lw $a0,0($sp) #fetch all the things j for_dfs1

 

用汇编语言写递归程序的经验教训:

1.值的进栈与出栈:这个一定要注意,每次调用其他过程的时候,都要想想是不是要把退路保存下来,把原来的变量保存下来,有些甚至不止保存一次,有些甚至会因为分支而要在各个分支中进栈出栈。递归的还得多练,这次居然写了六个小时还被卡了一个点,由此可见我P2大概率会被干掉。

2.判断条件的书写:很多时候用相等判断beq并不够(因为自己写的程序太混乱,值会超过预期),我们需要能判断大于等于,小于等于,严格小,严格大的指令,至于到底使用哪种,一定要在写代码的时候想好。循环退出的条件一般写在循环的最前面。最好每个条件判断后面都写上清晰的注释。

3.别怕。汇编程序看着长,实际上重复的部分有很多是进栈出栈,以后写的时候不能再抓狂到砸东西了!

posted @ 2019-10-20 00:28  BUAA-Wander  阅读(850)  评论(0编辑  收藏  举报