codeforces 821d Okabe and City

D. Okabe and City

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Okabe likes to be able to walk through his city on a path lit by street lamps. That way, he doesn't get beaten up by schoolchildren.

Okabe's city is represented by a 2D grid of cells. Rows are numbered from 1 to n from top to bottom, and columns are numbered 1 to mfrom left to right. Exactly k cells in the city are lit by a street lamp. It's guaranteed that the top-left cell is lit.

Okabe starts his walk from the top-left cell, and wants to reach the bottom-right cell. Of course, Okabe will only walk on lit cells, and he can only move to adjacent cells in the up, down, left, and right directions. However, Okabe can also temporarily light all the cells in any single row or column at a time if he pays 1 coin, allowing him to walk through some cells not lit initially.

Note that Okabe can only light a single row or column at a time, and has to pay a coin every time he lights a new row or column. To change the row or column that is temporarily lit, he must stand at a cell that is lit initially. Also, once he removes his temporary light from a row or column, all cells in that row/column not initially lit are now not lit.

Help Okabe find the minimum number of coins he needs to pay to complete his walk!

Input

The first line of input contains three space-separated integers n, m, and k (2 ≤ n, m, k ≤ 104).

Each of the next k lines contains two space-separated integers ri and ci (1 ≤ ri ≤ n, 1 ≤ ci ≤ m) — the row and the column of the i-th lit cell.

It is guaranteed that all k lit cells are distinct. It is guaranteed that the top-left cell is lit.

Output

Print the minimum number of coins Okabe needs to pay to complete his walk, or -1 if it's not possible.

Examples

input

4 4 51 12 12 33 34 3

output

2

input

5 5 41 12 13 13 2

output

-1

input

2 2 41 11 22 12 2

output

0

input

5 5 41 12 23 34 4

output

3

Note

In the first sample test, Okabe can take the path , paying only when moving to (2, 3) and (4, 4).

In the fourth sample, Okabe can take the path  , paying when moving to (1, 2), (3, 4), and (5, 4).

题意：有一个n*m地图，有k个位置是点亮的,有4个移动方向，每次可以移动到相邻的点亮位置,每次站在初始被点亮某个位置，暂时使某行或该某列全部点亮，花费为1，下一次使用时，上一次暂时点亮被熄灭。

思路：最短路: 以K个在初始就被点亮的点建图，如果有在四联通下相邻的灯则两点之间权为0，如果两点之间横/纵坐标相差不超过2，也就是说相隔一行/列，则两点之间权为1.

代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;const int maxn=10005;const int inf=1e9-1;int n,m,k;int u[maxn],v[maxn];int vis[maxn],d[maxn];int SPFA(int start){    memset(vis,0,sizeof(vis));    for(int i=0;i<maxn;i++)        d[i]=inf;    queue<int>que;    while(!que.empty()) que.pop();    que.push(start),d[start]=0;    vis[start]=1;    while(!que.empty())    {        int x=que.front();        que.pop();        vis[x]=0;        for(int i=1;i<=k;i++)        {            if(i==x) continue;            int w=inf;            int p=abs(u[i]-u[x]),q=abs(v[i]-v[x]);            if(p+q==1) w=0;            else if(p<=2||q<=2) w=1;            if(d[i]>d[x]+w)            {                d[i]=d[x]+w;                if(!vis[i])                {                    vis[i]=1;                    que.push(i);                }            }        }    }    return d[k]>=inf?-1:d[k];}int main(){    scanf("%d%d%d",&n,&m,&k);    memset(u,0,sizeof(u));    memset(v,0,sizeof(v));    int flag=0;    for(int i=1;i<=k;i++)    {        scanf("%d%d",&u[i],&v[i]);        if(u[i]==n&&v[i]==m)            flag=1;    }    if(!flag)        u[++k]=n+1,v[k]=m+1;    int ans=SPFA(1);    printf("%d\n",ans);}