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;
}