CF1301F - Super Jaber

发表于 | 阅读数

一张图两种方式

  • 随便走一步
  • 跳到相同颜色的点

询问从$(x1,y1)\rightarrow (x2,y2)$最少步数。$n,m\leq 1000,c\leq 40$

从某个点到某个点，坑定是从某个颜色跳到某个颜色。暴力枚举所有颜色。

预处理出每种颜色到不同点的距离，因为求距离最小值，暴力$BFS$即可。
$O(cnm+qc)$

代码 
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  </details>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e3 + 10;
const int mod = 1e9 + 7;
const int M = 45;
int dx[] = {0, 0, 1, -1};
int dy[] = {-1, 1, 0, 0};
int n, m;
int dis[M][N][N], vis[N][N], vc[M];
int k, c[N][N];
vector<pii> g[M];
void bfs(int col)
{
    //cout << col << "----" << endl;
    queue<pii> q;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            vis[i][j] = 0, dis[col][i][j] = 1e9;
    for (int i = 1; i <= k; i++)
        vc[i] = 0;
    for (pii s : g[col])
    {
        dis[col][s.first][s.second] = 0;
        vis[s.first][s.second] = 1;
        q.push(s);
    }

    vc[col] = 1;
    while (!q.empty())
    {
        pii now = q.front();
        int x = now.first;
        int y = now.second;
        q.pop();
        if (!vc[c[x][y]])
        {

            for (pii s : g[c[x][y]])
            {
                if (!vis[s.first][s.second])
                {
                    dis[col][s.first][s.second] = min(dis[col][s.first][s.second], dis[col][x][y] + 1);
                    q.push(mk(s.first, s.second));
                    vis[s.first][s.second] = 1;
                }
            }
            vc[c[x][y]] = 1;
        }
        for (int i = 0; i < 4; i++)
        {
            int xx = x + dx[i];
            int yy = y + dy[i];
            if (xx >= 1 && xx <= n && yy >= 1 && yy <= m && !vis[xx][yy])
            {
                dis[col][xx][yy] = min(dis[col][xx][yy], dis[col][x][y] + 1);
                vis[xx][yy] = 1;
                q.push(mk(xx, yy));
            }
        }
    }
}

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", &c[i][j]);
            g[c[i][j]].push_back(mk(i, j));
        }
    for (int i = 1; i <= k; i++)
        bfs(i);
    int q;
    scanf("%d", &q);
    for (int i = 1; i <= q; i++)
    {
        int x1, y1, x2, y2;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        int ans = abs(x1 - x2) + abs(y1 - y2);
        for (int j = 1; j <= k; j++)
            ans = min(ans, dis[j][x1][y1] + dis[j][x2][y2] + 1);

        printf("%d\n", ans);
    }
}