BFS_1

Description

Background 
Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fast. Can you beat him? 
The Problem 
Your task is to write a program to calculate the minimum number of moves needed for a knight to reach one point from another, so that you have the chance to be faster than Somurolov. 
For people not familiar with chess, the possible knight moves are shown in Figure 1. 

Input

The input begins with the number n of scenarios on a single line by itself. 
Next follow n scenarios. Each scenario consists of three lines containing integer numbers. The first line specifies the length l of a side of the chess board (4 <= l <= 300). The entire board has size l * l. The second and third line contain pair of integers {0, ..., l-1}*{0, ..., l-1} specifying the starting and ending position of the knight on the board. The integers are separated by a single blank. You can assume that the positions are valid positions on the chess board of that scenario.

Output

For each scenario of the input you have to calculate the minimal amount of knight moves which are necessary to move from the starting point to the ending point. If starting point and ending point are equal,distance is zero. The distance must be written on a single line.

Sample Input

380 07 01000 030 50101 11 1

Sample Output

5280

广搜题；

题目意思是国际象棋中的马以“日”字走法从起点走到终点，输出最短步数。这显然是bfs解决，只要在结构体里面添加一个计数变量就可以

#include<iostream>#include<cstdio>#include<cstring>#include<queue>using namespace std;const int Maxn = 600;struct point{  int x;  int y;  int dis;};int a[Maxn][Maxn];//定义标记数组/*8个前进方向*/int X[8]={-2,-2,-1,-1,2,2,1,1};int Y[8]={-1,1,-2,2,-1,1,-2,2};int n;queue<point>que;point start,end;//  定义起点和终点int bfs(){  start.dis=0;  // 注意： start.dis 要提前赋值为 0  que.push(start);  a[start.x][start.y]=1;  // 标记此点已被访问  point temp,next;  while(1)  {    temp=que.front();//访问队首元素，即最早被压入队列的元素。    que.pop();  //出队， 弹出队列的第一个元素，注意，并不会返回被弹出元素的值。    int k;    for(k=0;k<8;k++)    {      next.x=temp.x+X[k];      next.y=temp.y+Y[k];      if(next.x>=0&&next.y>=0&&next.x<n&&next.y<n&&a[next.x][next.y]==0)  //判断能否入队      {        next.dis=temp.dis+1;        if(next.x == end.x && next.y == end.y)        return next.dis;        que.push(next); //入队,将next接到队列的末端。        a[next.x][next.y]=1;      }    }  }}int main(){    int t;    scanf("%d",&t);    while (t--)    {        scanf("%d",&n);        memset(a,0,sizeof(a));        scanf("%d%d",&start.x,&start.y);        scanf("%d%d",&end.x,&end.y);        if(start.x == end.x && start.y == end.y)        {            printf("0\n");            continue;        }        while(!que.empty())//判断队列空，如例：q.empty()，当队列空时，返回true。        que.pop();        printf("%d\n",bfs());    }    return 0;}