Significance: There is an n*m map, k positions are lighted, there are four moving directions, each time can move to adjacent lighted positions, each time standing at the initial lighted a position, temporarily make a row or a column all lighted up, cost 1, the next time when used, the last temporary lighted was extinguished.

Algorithms: Shortest path (many people do it separately, but only violence...................................................... The weight between two points is 0 if there are adjacent lights under four links. If the horizontal/vertical coordinates between the two points are not more than 2, that is to say, one row/column apart, the weight between the two points is 1. Because direct linking leads to MLE, it is enough to handle the linking in spfa.

Hint: If (n,m) is not lit at the beginning, then adding a point (n + 1, m + 1) is understood as forcing the lighting of the end of the original map.

Code:

#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>

using namespace std;

int rd() {
    int x = 0; char c = getchar();
    while (c > '9' || c < '0') c = getchar();
    while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
    return x;
}

const int N = 1e4 + 10;
const int inf = 0x7f7f7f7f;
int dis[N], x[N], y[N], n, m, K;
bool fg, vis[N];

int spfa() {
    queue <int> q;
    memset(dis, 0x7f, sizeof dis);
    q.push(1), vis[1] = 1, dis[1] = 0;
    while (!q.empty()) {
        int u = q.front(); q.pop();
        for (int i = 1; i <= K; i ++) {
            if (i == u) continue;
            int w = inf;
            int dx = abs(x[i] - x[u]), dy = abs(y[i] - y[u]);
            if (dx + dy == 1) w = 0;
            else if (dx <= 2 || dy <= 2) w = 1;
            if (dis[i] > dis[u] + w) {
                dis[i] = dis[u] + w;
                if (!vis[i]) vis[i] = 1, q.push(i);
            } 
        }
        vis[u] = 0;
    }
    return dis[K] == inf ? -1 : dis[K];
}

int main() {
    n = rd(), m = rd(), K = rd();
    for (int i = 1; i <= K; i ++) {
        x[i] = rd(), y[i] = rd();
        fg |= (x[i] == n && y[i] == m);
    }
    if (!fg) x[++K] = n + 1, y[K] = m + 1;
    printf("%d\n", spfa());
    return 0;
}