搜索：zoj 1091 Knight Moves （广搜）
【转】http://blog.csdn.net/zxy_snow/article/details/5741881
#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;
}
