【JZOJ6293】迷宫

description

---

analysis

• 有没有想起【\(NOIP2018\)】保卫王国？

• 设\(tr[t][x][y]\)表示线段树上的\(t\)节点代表的区间，从最左边列的\(x\)行到最右边列\(y\)行的最小距离

• 当区间长度为\(1\)时预处理很简单，注意向上走和向下走

• 合并两个区间\(2t,2t+1\)成\(t\)时，枚举中转点\(z\)，\(tr[t][x][y]=min(tr[2t][x][z]+tr[2t+1][z][y]+1)\)

• 对其实这个很像弗洛伊德的\(DP\)，就是拿两个区间拼起来

• 查询就线段树上面合并出一个大区间，然后就可以直接知道答案

• 修改就修改某个叶子节点代表的区间，重新处理数据，再向上合并

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 200005
#define INF 1000000007
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll a[6][MAXN];
bool bz[6][MAXN];
ll n,m,q;

struct node
{
	ll f[6][6];
}tr[MAXN<<2],tmp;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline ll max(ll x,ll y){return x>y?x:y;}
inline ll min(ll x,ll y){return x<y?x:y;}
inline void clear(node &tmp,ll pos)
{
	fo(i,1,n)fo(j,1,n)tmp.f[i][j]=INF;
	fo(i,1,n)
	{
		if (bz[i][pos])tmp.f[i][i]=0;
		fd(j,i-1,1)if (bz[j][pos] && tmp.f[i][j+1]<INF)tmp.f[i][j]=tmp.f[i][j+1]+1;else break;
		fo(j,i+1,n)if (bz[j][pos] && tmp.f[i][j-1]<INF)tmp.f[i][j]=tmp.f[i][j-1]+1;else break;
	}
}
inline node merge(node a,node b)
{
	node c;
	fo(i,1,n)fo(j,1,n)c.f[i][j]=INF;
	fo(k,1,n)fo(i,1,n)fo(j,1,n)
	c.f[i][j]=min(c.f[i][j],a.f[i][k]+b.f[k][j]+1);
	return c;
}
inline void build(ll t,ll l,ll r)
{
	if (l==r)
	{
		clear(tr[t],l);
		return;
	}
	ll mid=(l+r)>>1;
	build(t<<1,l,mid),build((t<<1)+1,mid+1,r);
	tr[t]=merge(tr[t<<1],tr[(t<<1)+1]);
}
inline void modify(ll t,ll l,ll r,ll x,ll y)
{
	if (l==r)
	{
		clear(tr[t],l);
		return;
	}
	ll mid=(l+r)>>1;
	if (x<=mid)modify(t<<1,l,mid,x,y);
	else modify((t<<1)+1,mid+1,r,x,y);
	tr[t]=merge(tr[t<<1],tr[(t<<1)+1]);
}
inline node query(ll t,ll l,ll r,ll x,ll y)
{
	if (l==x && y==r)return tr[t];
	ll mid=(l+r)>>1;
	if (y<=mid)return query(t<<1,l,mid,x,y);
	else if (x>mid)return query((t<<1)+1,mid+1,r,x,y);
	else return merge(query(t<<1,l,mid,x,mid),query((t<<1)+1,mid+1,r,mid+1,y));
}
int main()
{
	//freopen("T1.in","r",stdin);
	freopen("maze.in","r",stdin);
	freopen("maze.out","w",stdout);
	n=read(),m=read(),q=read();
	fo(i,1,n)fo(j,1,m)bz[i][j]=(a[i][j]=read());
	build(1,1,m);
	while (q--)
	{
		ll opt=read(),x=read(),y=read(),xx,yy;
		if (opt==1)bz[x][y]^=1,modify(1,1,m,y,x);
		else
		{
			xx=read(),yy=read();
			if (!bz[x][y] || !bz[xx][yy]){printf("-1\n");continue;}
			tmp=query(1,1,m,y,yy);
			printf("%lld\n",tmp.f[x][xx]<INF?tmp.f[x][xx]:-1ll);
		}
	}
	return 0;
}