bzoj 1627: [Usaco2007 Dec]穿越泥地【bfs】

在洛谷上被卡了一个点开了O2才过= =

bfs即可，为方便存储，把所有坐标+500

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
const int N=1005,dx[]={-1,1,0,0},dy[]={0,0,-1,1};
int n,sx,sy;
bool a[N][N],v[N][N];
struct qwe
{
	int x,y,b;
	qwe(int X=0,int Y=0,int B=0)
	{
		x=X,y=Y,b=B;
	}
};
int read()
{
	int r=0,f=1;
	char p=getchar();
	while(p>'9'||p<'0')
	{
		if(p=='-')
			f=-1;
		p=getchar();
	}
	while(p>='0'&&p<='9')
	{
		r=r*10+p-48;
		p=getchar();
	}
	return r*f;
}
bool ok(int x,int y)
{
	return x>=0&&y>=0&&x<=1000&&y<=1000&&!v[x][y]&&!a[x][y];
}
int main()
{
	sx=read()+500,sy=read()+500,n=read();
	for(int i=1;i<=n;i++)
	{
		int x=read()+500,y=read()+500;
		a[x][y]=1;
	}
	queue<qwe>q;
	q.push(qwe(500,500,0));
	v[500][500]=1;
	while(!q.empty())
	{
		int x=q.front().x,y=q.front().y,b=q.front().b;
		if(x==sx&&y==sy)
		{
			printf("%d\n",b);
			break;
		}
		q.pop();
		v[x][y]=0;
		for(int i=0;i<4;i++)
			if(ok(x+dx[i],y+dy[i]))
				q.push(qwe(x+dx[i],y+dy[i],b+1)),v[x+dx[i]][y+dy[i]]=1;
	}
	return 0;
}