#include <bits/stdc++.h>
using namespace std;
int n, m, k, ans, h[3005][3005], u[9000005], v[9000005], disf[3005][3005], diswb[3005][3005], diswe[3005][3005], tmp1 = 0x3f3f3f3f, tmp2 = 0x3f3f3f3f;
pair <int, int> mdp1, mdp2;
const int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};
bool vis[3005][3005];
bool sensible(int x, int y) {
    return x >= 1 && x <= n && y >= 1 && y <= m;
}
void bfswb()  {
    queue <pair <int, int> > q;
    pair <int, int> x;
    memset(diswb, 0x3f3f3f3f, sizeof(diswb));
    memset(vis, 0, sizeof(vis));
    q.push({1, 1}), diswb[1][1] = 0, vis[1][1] = 1;
    while(q.size()) {
        x = q.front(), q.pop();
        for(int i = 0; i < 4; i++)  {
            if(!vis[x.first + dx[i]][x.second + dy[i]] && h[x.first + dx[i]][x.second + dy[i]] != 0 && sensible(x.first + dx[i], x.second + dy[i])) {
                diswb[x.first + dx[i]][x.second + dy[i]] = min(diswb[x.first][x.second] + 1, diswb[x.first + dx[i]][x.second + dy[i]]);
                q.push({x.first + dx[i], x.second + dy[i]});
                vis[x.first + dx[i]][x.second + dy[i]] = 1;
            }
        }
    }
}
void bfswe()  {
    queue <pair <int, int> > q;
    pair <int, int> x;
    memset(diswe, 0x3f3f3f3f, sizeof(diswe));
    memset(vis, 0, sizeof(vis));
    q.push({n, m}), diswe[n][m] = 0, vis[1][1] = 1;;
    while(q.size()) {
        x = q.front(), q.pop();

        for(int i = 0; i < 4; i++)  {
            if(!vis[x.first + dx[i]][x.second + dy[i]] && h[x.first + dx[i]][x.second + dy[i]] != 0 && sensible(x.first + dx[i], x.second + dy[i])) {
                diswe[x.first + dx[i]][x.second + dy[i]] = min(diswe[x.first][x.second] + 1, diswe[x.first + dx[i]][x.second + dy[i]]);
                q.push({x.first + dx[i], x.second + dy[i]});
                vis[x.first + dx[i]][x.second + dy[i]] = 1;
            }
        }
    }
}
int main()  {
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            scanf("%d", &h[i][j]);
        }
    }
    for(int i = 1; i <= k; i++)
        scanf("%d%d", &u[i], &v[i]);
    bfswb(), bfswe();

    ans = diswb[n][m];
    for(int i = 1; i <= k; i++)  {
        if(diswb[u[i]][v[i]] <= tmp1)   {
            mdp1.first = u[i], mdp1.second = v[i], tmp1 = diswb[u[i]][v[i]];
        }
    }
    for(int i = 1; i <= k; i++)  {
        if(diswe[u[i]][v[i]] <= tmp2)    {
            mdp2.first = u[i], mdp2.second = v[i], tmp2 = diswe[u[i]][v[i]];
        }
    }
    ans = min(ans, tmp1 + tmp2 + (h[mdp1.first][mdp1.second] != h[mdp2.first][mdp2.second]) + 1);
    if(ans >= 0x3f3f3f3f)   ans = -1;
    printf("%d\n", ans);
    return 0;
}

大概思路：使用仙法显然至多一次，答案为两点之间最短路（若存在）和 从起点到御剑飞行点 1 距离、御剑飞行点 2 到终点距离与两飞行点的传递距离（1 或 2）之和（若存在），若都不存在即为 -1。

希望有人解答，感谢！