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

来源：互联网发布：java小程序代码编辑：程序博客网时间：2024/06/06 02:15

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;  }