zoj 1091 Knight Moves （BFS）（情况用循环控制，值得学习啊）

刚开始看不懂题，后来查了查才明白这颗棋子是走“日”字的。。。。

接着，那就会出现8种情况。。。无语，不过，有了2165的模板，我居然慢悠悠的把这八种情况罗列出来并且AC了。。

上一篇2165就想把这种相同部分写成函数来调用，可是又发现，相同部分里，if语句中有break；这怎么写到函数里啊？？？？？如果仅仅是把else部分（就两句话）写成函数，感觉又没那个必要了。。。纠结中。。

#include<stdio.h>
#include<string.h>
typedef struct
{
int s1[5000];
int s2[5000];
}queue;
queue que;
int head,rear;
void In(int x,int y)
{
que.s1[rear]=x;
que.s2[rear]=y;
rear++;
}
void Out(int *x,int *y)
{
*x=que.s1[head];
*y=que.s2[head];
head++;
}

int isEmpty()
{
if(head==rear)
return 1;
else
return 0;
}
int main()
{
int w,h,i,j,x1,x2,y1,y2,count[9][9],b[9][9],p1,p2;
char temp,t1,t2;
while(scanf("%c%d %c%d",&t1,&y1,&t2,&y2)!=EOF)
{
x1=t1-'a'+1;
x2=t2-'a'+1;
for(i=0;i<9;i++)
for(j=0;j<9;j++)
{
b[i][j]=0;
count[i][j]=0;
}
head=rear=0;
In(x1,y1);
b[x1][y1]=1;
while(!isEmpty())
{
Out(&p1,&p2);
if(p1==x2&&p2==y2)
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",t1,y1,t2,y2,count[p1][p2]);
break;
}
else
{

i=p1-2,j=p2+1;
if(i>=1&&j<=8&&b[i][j]==0)
{

count[i][j]=count[p1][p2]+1;
if(i==x2&&j==y2)
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",t1,y1,t2,y2,count[i][j]);
break;
}
else
{
In(i,j);
b[i][j]=1;
}
}

i=p1-1,j=p2+2;
if(i>=1&&j<=8&&b[i][j]==0)
{
count[i][j]=count[p1][p2]+1;
if(i==x2&&j==y2)
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",t1,y1,t2,y2,count[i][j]);
break;
}
else
{
In(i,j);
b[i][j]=1;
}
}

i=p1+1,j=p2+2;
if(i<=8&&j<=8&&b[i][j]==0)
{
count[i][j]=count[p1][p2]+1;
if(i==x2&&j==y2)
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",t1,y1,t2,y2,count[i][j]);
break;
}
else
{
In(i,j);
b[i][j]=1;
}
}
i=p1+2,j=p2+1;
if(i<=8&&j<=8&&b[i][j]==0)
{
count[i][j]=count[p1][p2]+1;
if(i==x2&&j==y2)
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",t1,y1,t2,y2,count[i][j]);
break;
}
else
{
In(i,j);
b[i][j]=1;
}
}
i=p1+2,j=p2-1;
if(i<=8&&j>=1&&b[i][j]==0)
{
count[i][j]=count[p1][p2]+1;
if(i==x2&&j==y2)
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",t1,y1,t2,y2,count[i][j]);
break;
}
else
{
In(i,j);
b[i][j]=1;
}
}
i=p1+1,j=p2-2;
if(i<=8&&j>=1&&b[i][j]==0)
{
count[i][j]=count[p1][p2]+1;
if(i==x2&&j==y2)
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",t1,y1,t2,y2,count[i][j]);
break;
}
else
{
In(i,j);
b[i][j]=1;
}
}
i=p1-1,j=p2-2;
if(i>=1&&j>=1&&b[i][j]==0)
{
count[i][j]=count[p1][p2]+1;
if(i==x2&&j==y2)
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",t1,y1,t2,y2,count[i][j]);
break;
}
else
{
In(i,j);
b[i][j]=1;
}
}
i=p1-2,j=p2-1;
if(i>=1&&j>=1&&b[i][j]==0)
{
count[i][j]=count[p1][p2]+1;
if(i==x2&&j==y2)
{
printf("To get from %c%d to %c%d takes %d knight moves.\n",t1,y1,t2,y2,count[i][j]);
break;
}
else
{
In(i,j);
b[i][j]=1;
}
}
}
}
getchar();
}
return 0;
}

我晕啊！！！！！！！弄个循环就把这8种情况搞定了！我靠lai！！！！这么先进，这么先进

。。。。。。忍了。。。。学习一下吧。。。。主要是那个循环控制8种情况，太纠结了。。。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int state[9][9];
int count[9][9];
int Queue[100000];
int step[8][2] = {1,2, 1,-2, -1,2, -1,-2, 2,1, 2,-1, -2,1, -2,-1};
int head,tail;
int push(int x)
{
Queue[head++] = x;
}
int pop(void)
{
return Queue[tail++];
}
int Qempty(void)
{
if( head == tail )
return 1;
return 0;
}
void init(void)
{
head = 0; tail = 0;
memset( state,0,sizeof(state) );
memset( count,0,sizeof(count) );
memset( Queue,0,sizeof(Queue) );
}
int main(void)
{
int a,b,x,y,temp,tempa,tempx,ta,tx,i;
char ch1,ch2,n;
while( scanf("%c%d %c%d%c",&ch1,&x,&ch2,&y,&n)!=EOF ) //这点很纠结，因为有个回车，不再输入一个回车的话，会错
{
init();
a = ch1 - 'a' + 1;
b = ch2 - 'a' + 1;
push(a); push(x);
state[a][x] = 1;
while( !Qempty() )
{
tempa = pop();
tempx = pop();
if( tempa == b && tempx == y)
break;
for(i=0; i<8; i++)//8个方向，用循环一一调用，这点很值得学习！
{
ta = tempa + step[i][0];
tx = tempx + step[i][1];
if(state[ta][tx] == 0 && ta>=1 && ta<=8 && tx>=1 && tx<=8 )
{
push(ta); push(tx);
state[ta][tx] = 1;
count[ta][tx] = count[tempa][tempx] + 1;
}
}
}
printf("To get from %c%d to %c%d takes %d knight moves./n",ch1,x,ch2,y,count[tempa][tempx]);
}
system("pause");
return 0;
}