poj1915 2.txt

1位全加器 可以进行1位的二进制码的加法 想进行改进 改为4位或8位的全加器代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct queue
{
int x,y;
int ste;
struct queue *next;
};

struct root
{
struct queue *pre;
struct queue *aft; 
}root;

int t;
int n;
int stx,sty,enx,eny;
int flag,fail ;
int ans;
int d[8][2] = {{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1},{-2,1},{-1,2}};
int mark[302][302];

void solve();

void bfs(int x,int y,int ste);

int main()
{
scanf("%d",&t);
while(t--)
{
   scanf("%d%d%d%d%d",&n,&stx,&sty,&enx,&eny);
   solve();
}
return 0;
}

void solve()
{
struct queue *p;
flag = 0;
fail = 0;
memset(mark,0,sizeof(mark));
p = (struct queue *)malloc(sizeof(struct queue));
p->x = stx;
p->y = sty;
p->ste = 0;
p->next = NULL; 
root.pre = p;
root.aft = p;
bfs(stx,sty,0);
printf("%d\n",ans);
}

void bfs(int x,int y,int ste)
{
int i;
struct queue *p;
if(flag || fail)
   return ;
if(x == enx && y == eny)
{
   ans = ste;
   flag = 1;
   return ;
}
for(i = 0;i < 8; i++)
{
   if( (x+d[i][0]>=0 && x+d[i][0] <n) && (y+d[i][1]>=0 && y+d[i][1]<n) && mark[x+d[i][0]][y+d[i][1]] == 0)
   {
    p = (struct queue *)malloc(sizeof(struct queue));
    p->x = x+d[i][0];
    p->y = y+d[i][1];
    p->ste = ste+1;
    p->next = NULL;
    root.aft->next = p;
    root.aft = p;
    mark[x+d[i][0]][y+d[i][1]] = 1;
   }
}
p = root.pre;
root.pre = p->next;
p->next = NULL;
free(p);
if(root.pre == NULL)
{
   fail = 1;
   return ;
}
bfs(root.pre->x,root.pre->y,root.pre->ste);
}