nyoj-58-最小步数

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#define L 9
using namespace std;
bool map[9][9]={{1,1,1,1,1,1,1,1,1},
{1,0,0,1,0,0,1,0,1},
{1,0,0,1,1,0,0,0,1},
{1,0,1,0,1,1,0,1,1},
{1,0,0,0,0,1,0,0,1},
{1,1,0,1,0,1,0,0,1},
{1,1,0,1,0,1,0,0,1},
{1,1,0,1,0,0,0,0,1},
{1,1,1,1,1,1,1,1,1}};
int num[9][9],dx[4]={1,0,-1,0},dy[4]={0,1,0,-1},dis[81],temp[9][9];//num数组存放到某一点的最小步数，dis为队列,temp标记是否访问过
int bfs(int x1,int y1,int x2,int y2);
int main()
{
int n,a,b,c,d;
scanf("%d",&n);
while(n--)
{
scanf("%d%d%d%d",&a,&b,&c,&d);
printf("%d\n",bfs(a,b,c,d));
}
return 0;
}

int bfs(int x1,int y1,int x2,int y2)
{
memset(num,0,sizeof(num));
memset(temp,0,sizeof(temp));
int l=0,r=0,t;
t=x1*L+y1;
dis[r++]=t;
temp[x1][y1]=1;
num[x1][y1]=0;
while(l<r)
{
t=dis[l++];
int x,y,bx,by;
x=t/L;y=t%L;
for(int i=0;i<4;i++)
{
bx=x+dx[i];by=y+dy[i];
if(temp[bx][by]==0&&bx>=0&&bx<L&&by>=0&&by<L&&map[bx][by]==0)
{
dis[r++]=bx*L+by;
num[bx][by]=num[x][y]+1;
temp[bx][by]=1;
}
}
}
return num[x2][y2];
}