COJ 1224 ACM Group's Wacky Chess

Be sure to pay attention to the input and output later

For example, the least-given coordinate of the topic starts from (

Then decide if the array is out of bounds.

We need to pay special attention.

I know the truth.

And then because of this problem and when you subtract 1 from the input coordinates,

I counted the m,n in the accident.

And then you blew me up for one hours.

Mom, a chicken.

That's the first way to play rank today.

I thought about it when I got the title.

But did not consider the column

Fried 2h.

I've got a dog's heart in my mother's Day.

#include <stdio.h> #include <queue> #include <string.h> #include <stdlib.h>//#include < Iostream>using namespace Std;int n,m,sx,sy,ex,ey;int map[21][21];int dir[8][2]={{-1,2},{1,2},//right jump 0,1 { 2,1},{2,-1},//jump up 2,3 {1,-2},{-1,-2},//left jump 4,5 { -2,-1},{-2,1}//jump down 6,7};struct node{int x,y,step;};  int check (int x,int y) {if (x>=0&&x<m&&y>=0&&y<n&&map[x][y]==0) return 1; else return 0;}    int BFS (int x,int y) {Node temp, next;    Queue<node> Queue;    Temp.x=x;    Temp.y=y;    Temp.step=0;    Map[x][y]=1;    Queue.push (temp); while (!        Queue.empty ()) {Temp=queue.front ();        Queue.pop ();        if (Temp.x==ex&&temp.y==ey) {return temp.step;                } else if (Temp.x==ex&&temp.y+1==ey)//up {for (int i=0; i<8;i++) {if (i==2| | i==3) Continue;                    NEXT.X=TEMP.X+DIR[I][0];                   NEXT.Y=TEMP.Y+DIR[I][1];                    if (check (NEXT.X,NEXT.Y) ==1) {next.step=temp.step+1;                    Map[next.x][next.y]=1;                    if (Next.x==ex&&next.y==ey) return next.step;                Queue.push (next);                }}} and else if (Temp.x==ex&&temp.y-1==ey)//down {for (int i=0; i<8;i++) {if (i==6| |                I==7) continue;                NEXT.X=TEMP.X+DIR[I][0];                NEXT.Y=TEMP.Y+DIR[I][1];                if (check (NEXT.X,NEXT.Y) ==1) {next.step=temp.step+1;                Map[next.x][next.y]=1;                if (Next.x==ex&&next.y==ey) return next.step;            Queue.push (next); }}} and else if (Temp.x+1==ex&&temp.y==ey)//right {for (int i=0; i<8;i++) {if (i==0| |                I==1) continue;                NEXT.X=TEMP.X+DIR[I][0];                NEXT.Y=TEMP.Y+DIR[I][1];                if (check (NEXT.X,NEXT.Y) ==1) {next.step=temp.step+1;                Map[next.x][next.y]=1;                if (Next.x==ex&&next.y==ey) return next.step;            Queue.push (next);                }}} and else if (Temp.x-1==ex&&temp.y==ey)//left {for (int i=0; i<8;i++) {if (i==4| |                I==5) continue;                NEXT.X=TEMP.X+DIR[I][0];                NEXT.Y=TEMP.Y+DIR[I][1];                if (check (NEXT.X,NEXT.Y) ==1) {next.step=temp.step+1;                Map[next.x][next.y]=1;                if (Next.x==ex&&next.y==ey) return next.step;            Queue.push (next); }}}} else for (int i=0; i<8;i++) {Next.x=teMP.X+DIR[I][0];            NEXT.Y=TEMP.Y+DIR[I][1];                if (check (NEXT.X,NEXT.Y) ==1) {next.step=temp.step+1;                Map[next.x][next.y]=1;            Queue.push (next); }}} return-1;}    int main () {//int ex,ey,sx,sy; while (scanf ("%d%d%d%d%d", &m,&n,&sx,&sy,&ex,&ey)!=eof) {memset (map,0,sizeof (MAP))        ;        sx--;sy--;ex--;ey--;    printf ("%d\n", BFS (Sx,sy)); } return 0;}

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COJ 1224 ACM Group's Wacky Chess