最少步数 南工acm58

先把距离为1可以到的位置存起来，然后由距离为一的位置找距离为2的位置，再由距离为2的位置找距离为3的位置存起来。。。

每次找完位置后，判断要到的地方在不在这次找到的距离为d的位置里，如果在，就结束，输出d，不在，就继续找！

# include <iostream>
# include <vector>
# include <string.h>
using namespace std;
bool map[9][9]={
	{1,1,1,1,1,1,1,1,1},
	{1,0,0,1,0,0,1,0,1},
	{1,0,0,1,1,0,0,0,1},
	{1,0,1,0,1,1,0,1,1},
	{1,0,0,0,0,1,0,0,1},
	{1,1,0,1,0,1,0,0,1},
	{1,1,0,1,0,1,0,0,1},
	{1,1,0,1,0,0,0,0,1},
	{1,1,1,1,1,1,1,1,1}
};
bool m[9][9];//标志数组
struct data
{
	int x, y;
};
vector<data> s[64];//存放距离分别为1，2，3...的位置
int main()
{
	int n, flag, i, j, k;
	data a, b, t;
	cin>>n;
	while (n--)
	{
		memset(m, 0, sizeof(m));
		memset(s, 0, sizeof(s));
		cin>>a.x>>a.y>>b.x>>b.y;
		s[0].push_back(a);
		flag = 0;
		if (a.x==b.x&&a.y==b.y)
		{	cout<<"0\n";continue;}
		for (i = 0; i<64; i++)
		{
			for (j = 0; j<s[i].size(); j++)
			{
				if (!m[s[i][j].x+1][s[i][j].y] && !map[s[i][j].x+1][s[i][j].y])//判断是否已经找过该点，和该点是不是空地，不是空地就继续找下一个距离的点
				{t.x = s[i][j].x+1; t.y = s[i][j].y; s[i+1].push_back(t);m[s[i][j].x+1][s[i][j].y]=1;}
				if (!m[s[i][j].x-1][s[i][j].y] && !map[s[i][j].x-1][s[i][j].y])
				{t.x = s[i][j].x-1; t.y = s[i][j].y; s[i+1].push_back(t);m[s[i][j].x-1][s[i][j].y]=1;}
				if (!m[s[i][j].x][s[i][j].y+1] && !map[s[i][j].x][s[i][j].y+1])
				{t.x = s[i][j].x; t.y = s[i][j].y+1; s[i+1].push_back(t);m[s[i][j].x][s[i][j].y+1]=1;}
				if (!m[s[i][j].x][s[i][j].y-1] && !map[s[i][j].x][s[i][j].y-1])
				{t.x = s[i][j].x; t.y = s[i][j].y-1; s[i+1].push_back(t);m[s[i][j].x][s[i][j].y-1]=1;}
			}
			for (k = 0; k<s[i+1].size(); k++)//判断要到的位置是不是已经找到
			{
				if (s[i+1][k].x == b.x&&s[i+1][k].y==b.y)
					flag = 1;
			}
			if (flag)//找到该点，flag=1，跳出循环，并输出步数
				break;
		}
		cout<<i+1<<endl;
	}
	return 0;
}