【CodeForces - 821D Okabe and City】 最短路spfa

E - Okabe and City

 

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 nfrom top to bottom, and columns are numbered 1 to m from left to right. Exactly kcells 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 1coin, 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 r_i and c_i (1 ≤ r_i ≤ n,1 ≤ c_i ≤ 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.

Example

Input

4 4 5
1 1
2 1
2 3
3 3
4 3

Output

2

Input

5 5 4
1 1
2 1
3 1
3 2

Output

-1

Input

2 2 4
1 1
1 2
2 1
2 2

Output

0

Input

5 5 4
1 1
2 2
3 3
4 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).

題意：給你一個,n*m的二維平面，其中有k個格子被路燈點亮，現在Okabe要從點(1,1)走到點(n,m)，他只能走被點亮的格子，而Okabe自己可以點亮一行或者一列的所有格子（在經過後會熄滅），問他最少多少次點亮格子可以走到點(n,m)。

分析:將所有被點亮的格子記錄，如果兩個格子相鄰，則把兩點間的邊權設爲0，如果兩個格子行或列之差爲1但不相鄰，則把兩點間的邊權設爲1，最後計算最短路就行了。需要注意的是，如果終點沒有被點亮，那麼需要設點(n+1,m+1)爲終點，並加入到點亮的點集合中。

代碼如下：

#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>

#define LL long long
#define INF 0x3f3f3f3f

using namespace std;
const int MX = 1e4 + 5;
int n, m, k;
struct node{
    int x, y;
}lit[MX];
int dis[MX];
int inq[MX];

int spfa(){
    for(int i = 1; i <= k; i++){
        dis[i] = INF;
    }
    queue<int> q;
    q.push(1);
    inq[1] = 1;
    dis[1] = 0;
    while(!q.empty()){
        int u = q.front();
        q.pop();
        inq[u] = 0;
        for(int i = 1; i <= k; i++){
            if(u == i)  continue;
            int val = INF;
            int dx = abs(lit[u].x - lit[i].x);
            int dy = abs(lit[u].y - lit[i].y);
            if(dx+dy == 1){
                val = 0;
            }
            else if(dx <= 2 || dy <= 2){
                val = 1;
            }
            if(dis[i] > dis[u] + val){
                dis[i] = dis[u] + val;
                if(!inq[i]){
                    q.push(i);
                    inq[i] = 1;
                }
            }
        }
    }
    if(dis[k] != INF)   return dis[k];
    return -1;
}


int main(){
    scanf("%d%d%d", &n, &m, &k);
    int flag = 0;
    for(int i = 1; i <= k; i++){
        int u, v;
        scanf("%d%d", &lit[i].x, &lit[i].y);
        if(lit[i].x == n && lit[i].y == m)  flag = 1;
    }
    if(!flag){
        lit[++k].x = n+1;
        lit[k].y = m+1;
    }
    cout << spfa() << endl;
    return 0;
}