CF1301F Super Jaber

https://www.luogu.com.cn/problem/CF1301F

因为$k$很小，所以我们可以记录$f[x][y][k]$表示从$(x,y)$出发，走到颜色为$k$的最短距离

建反图跑bfs即可

然后简单分类讨论一下即可

写到一半确切感觉到自己基础不扎实

#include<bits/stdc++.h>
#define N 1000050
using namespace std;
inline int read() {
    int x = 0; char ch = getchar();
    for(; ch < '0' || ch > '9'; ) ch = getchar();
    for(; ch >= '0' && ch <= '9'; ) x = (x << 3) + (x <<1) + (ch - '0'), ch = getchar();
    return x;
}
vector<int> g[N], a[N];
void insert(int u, int v) {
    g[u].push_back(v), g[v].push_back(u);
}
int n, m, c, qq;
short dis[42][N], col[N], visc[42];
inline int id(int x, int y) {
    return (x - 1) * m + y;
}
queue<int> q;
void dfs(int X) {
    for(int i = 1; i <= n * m; i ++) 
        dis[X][i] = -1;
    
    for(int i = 1; i <= c; i ++) visc[i] = 0;
    visc[X] = 1;
    for(int x : a[X]) dis[X][x] = 0, q.push(x);

    int gs = 0;
    while(q.size()) {
        int u = q.front(); q.pop();
        if(!visc[col[u]])
        for(int v : a[col[u]]) if(dis[X][v] == -1) {
            dis[X][v] = dis[X][u] + 1;
            q.push(v);
        } 
        visc[col[u]] = 1;

        for(int v : g[u]) if(dis[X][v] == -1) {
            dis[X][v] = dis[X][u] + 1;
            q.push(v);
        }
    }
}
int main() {
    n = read(), m = read(), c = read();
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++) {
            int x;
            x = read();
            col[id(i, j)] = x;
            a[x].push_back(id(i, j));
        }
    
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m; j ++) {
            if(i > 1) insert(id(i - 1, j), id(i, j));
            if(j > 1) insert(id(i, j - 1), id(i, j));
        }

    for(int k = 1; k <= c; k ++) if(a[k].size()) {
        dfs(k);
    }

    qq = read();
    while(qq --) {
        int x, y, xx, yy;
        x = read(), y = read(), xx = read(), yy = read();
        int ans = abs(x - xx) + abs(y - yy);
        for(int k = 1; k <= c; k ++) if(a[k].size())  ans = min(ans, dis[k][id(x, y)] + dis[k][id(xx, yy)] + 1);
        printf("%d\n", ans);       
    }
    return 0;
}