样例4答案是3 我的代码输出2

#include<iostream>
#include<queue>
using namespace std;
int n,m,k,mynext[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
int a[3010][3010],ans[3010][3010];
bool v[3010][3010],y[3010][3010];
struct type{
	int x,y;
}t[9000010];
queue<type> q;
inline int read()
{
	int x=0,f=1;
	char ch=getchar();
	while(ch<'0' || ch>'9')
	{
		if(ch=='-') f=-1;
		ch=getchar();
	}
	while(ch>='0' && ch<='9')
	{
		x=x*10+ch-'0';
		ch=getchar(); 
	}
	return f*x;
}
void bfs()
{
	v[1][1]=1;
	q.push((type){1,1});
	while(!q.empty())
	{
		type now=q.front();
		q.pop();
		if(now.x==n&&now.y==m) return ;
		for(int i=0;i<4;i++)
		{
			int newx=now.x+mynext[i][0],newy=now.y+mynext[i][1];
			if(newx>n||newx<1||newy>m||newy<1||a[newx][newy]==0||v[newx][newy]==1) continue;
			v[newx][newy]=1;
			ans[newx][newy]=ans[now.x][now.y]+1;
			q.push(type{newx,newy});
		}
		if(y[now.x][now.y]==1)
		{
			for(int i=1;i<=k;i++)
			{
				if(t[i].x!=now.x&&t[i].x!=now.y)
				{
					if(a[t[i].x][t[i].y]!=0)
					{
						v[t[i].x][t[i].y]=1;
						if(v[t[i].x][t[i].y]!=0) ans[t[i].x][t[i].y]=min(ans[now.x][now.y]+1,ans[t[i].x][t[i].y]);
						else ans[t[i].x][t[i].y]=ans[now.x][now.y]+1;
						q.push((type){t[i].x,t[i].y});
					}
				}
			}
		}
	}
}
int main()
{
	cin>>n>>m>>k;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			a[i][j]=read();
		}
	}
	for(int i=1;i<=k;i++)
	{
		int x1,y1;
		x1=read(),y1=read();
		y[x1][y1]=1;
		t[i]=(type){x1,y1};
	}
	bfs();
	if(ans[n][m]!=0) cout<<ans[n][m];
	else cout<<-1;
	return 0;
}

悬赏1关注，lz可能只有晚上能看