cf413E Maze 2D

E. Maze 2D

time limit per test 2 seconds

memory limit per test 256 megabytes

input standard input

output standard output

The last product of the R2 company in the 2D games' field is a new revolutionary algorithm of searching for the shortest path in a 2 × nmaze.

Imagine a maze that looks like a 2 × n rectangle, divided into unit squares. Each unit square is either an empty cell or an obstacle. In one unit of time, a person can move from an empty cell of the maze to any side-adjacent empty cell. The shortest path problem is formulated as follows. Given two free maze cells, you need to determine the minimum time required to go from one cell to the other.

Unfortunately, the developed algorithm works well for only one request for finding the shortest path, in practice such requests occur quite often. You, as the chief R2 programmer, are commissioned to optimize the algorithm to find the shortest path. Write a program that will effectively respond to multiple requests to find the shortest path in a 2 × n maze.

Input

The first line contains two integers, n and m (1 ≤ n ≤ 2·105; 1 ≤ m ≤ 2·105) — the width of the maze and the number of queries, correspondingly. Next two lines contain the maze. Each line contains n characters, each character equals either '.' (empty cell), or 'X' (obstacle).

Each of the next m lines contains two integers vi and ui (1 ≤ vi, ui ≤ 2n) — the description of the i-th request. Numbers vi, ui mean that you need to print the value of the shortest path from the cell of the maze number vi to the cell number ui. We assume that the cells of the first line of the maze are numbered from 1 to n, from left to right, and the cells of the second line are numbered from n + 1 to 2n from left to right. It is guaranteed that both given cells are empty.

Output

Print m lines. In the i-th line print the answer to the i-th request — either the size of the shortest path or -1, if we can't reach the second cell from the first one.

Sample test(s)

input

4 7
.X..
...X
5 1
1 3
7 7
1 4
6 1
4 7
5 7

output

1
4
0
5
2
2
2

input

10 3
X...X..X..
..X...X..X
11 7
7 18
18 10

output

9
-1
3

这……线段树神题啊

但是这是道馆之战的弱化版……

道馆之战是树上的情况，这题只是一条链的情况

用线段树维护每一个1*2的格子从第一格能不能到第一格、从第一格能不能到第二格、从第二格能不能到第一格、从第二格能不能到第二格

#include<cstdio>
#include<iostream>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
    LL 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 x*f;
}
struct segtree{
	int l,r;
	int a_to_a,a_to_b,b_to_a,b_to_b;
}tree[1000010];
segtree query;
bool mrk[2][200010];
int n,m,x0,y0,x1,y1;
segtree merge(segtree a,segtree b)
{
	segtree k;
	k.a_to_a=k.a_to_b=k.b_to_a=k.b_to_b=-1;
	k.l=min(a.l,b.l); k.r=max(a.r,b.r);

	if (a.a_to_a!=-1&&b.a_to_a!=-1)k.a_to_a=a.a_to_a+b.a_to_a+1;
	if (a.a_to_b!=-1&&b.b_to_a!=-1)
	{
		if (k.a_to_a==-1)k.a_to_a=a.a_to_b+b.b_to_a+1;
		else k.a_to_a=min(k.a_to_a,a.a_to_b+b.b_to_a+1);
	}

	if (a.a_to_a!=-1&&b.a_to_b!=-1)k.a_to_b=a.a_to_a+b.a_to_b+1;
	if (a.a_to_b!=-1&&b.b_to_b!=-1)
	{
		if (k.a_to_b==-1)k.a_to_b=a.a_to_b+b.b_to_b+1;
		else k.a_to_b=min(k.a_to_b,a.a_to_b+b.b_to_b+1);
	}

	if (a.b_to_a!=-1&&b.a_to_a!=-1)k.b_to_a=a.b_to_a+b.a_to_a+1;
	if (a.b_to_b!=-1&&b.b_to_a!=-1)
	{
		if (k.b_to_a==-1)k.b_to_a=a.b_to_b+b.b_to_a+1;
		else k.b_to_a=min(k.b_to_a,a.b_to_b+b.b_to_a+1);
	}

	if (a.b_to_a!=-1&&b.a_to_b!=-1)k.b_to_b=a.b_to_a+b.a_to_b+1;
	if (a.b_to_b!=-1&&b.b_to_b!=-1)
	{
		if (k.b_to_b==-1)k.b_to_b=a.b_to_b+b.b_to_b+1;
		else k.b_to_b=min(k.b_to_b,a.b_to_b+b.b_to_b+1);
	}
	return k;
}
inline void buildtree(int now,int l,int r)
{
	tree[now].l=l;tree[now].r=r;
	if (l==r)
	{
		tree[now].a_to_a=tree[now].a_to_b=tree[now].b_to_a=tree[now].b_to_b=-1;
		if (mrk[0][l])tree[now].a_to_a=0;
		if (mrk[1][l])tree[now].b_to_b=0;
		if (mrk[0][l]&&mrk[1][l])
		{
			tree[now].a_to_b=1;
			tree[now].b_to_a=1;
		}
		return;
	}
	int mid=(l+r)>>1;
	buildtree(now<<1,l,mid);
	buildtree(now<<1|1,mid+1,r);
	tree[now]=merge(tree[now<<1],tree[now<<1|1]);
}
inline void work(int now,int x,int y)
{
	int l=tree[now].l,r=tree[now].r;
	if (l==x&&r==y)
	{
		if (!query.l)query=tree[now];
		else query=merge(query,tree[now]);
		return;
	}
	int mid=(l+r)>>1;
	if (y<=mid)work(now<<1,x,y);
	else if (x>mid)work(now<<1|1,x,y);
	else
	{
		work(now<<1,x,mid);
		work(now<<1|1,mid+1,y);
	}
}
inline int ask(int x0,int y0,int x1,int y1)
{
	query.l=query.r=0;
	work(1,y0,y1);
	if (!x0&&!x1)return query.a_to_a;
	if (!x0&&x1)return query.a_to_b;
	if (x0&&!x1)return query.b_to_a;
	if (x0&&x1)return query.b_to_b;
}
int main()
{
	n=read();m=read();
	for(int i=0;i<=1;i++)
	for(int j=1;j<=n;j++)
	{
		char ch=getchar();while (ch!='X'&&ch!='.')ch=getchar();
		if (ch=='.')mrk[i][j]=1;
	}
	buildtree(1,1,n);
	for(int i=1;i<=m;i++)
	{
		y0=read();y1=read();x0=x1=0;
		if ((y0-1)%n+1>(y1-1)%n+1)swap(y0,y1);
		if (y0>n){x0=1;y0-=n;}
		if (y1>n){x1=1;y1-=n;}
		printf("%d\n",ask(x0,y0,x1,y1));
	}
}