2018ICPC焦作F-Honeycomb 蜂巢BFS

Posted 2021-12-12 hesorchen

题目

https://codeforces.com/gym/102028/problem/F

求S-T至少要经过几个蜂巢，例如下图为7。

题解

注意细节，按照题意模拟即可。

代码

#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <iostream>
using namespace std;

const int N = 1e3 + 10;
char s[4 * N][6 * N];
int id[4 * N][6 * N];

int mov[][2] = {2, 0, -1, 3, 1, 3};
int sid, eid;

vector<int> mp[N * N];

void add(int u, int v)
{
    mp[u].emplace_back(v);
    mp[v].emplace_back(u);
}

bool vis[N * N];
int bfs()
{
    queue<pair<int, int>> q;
    q.push(make_pair(sid, 0));
    vis[sid] = 1;
    while (q.size())
    {
        int u = q.front().first;
        int step = q.front().second;
        q.pop();
        if (u == eid)
        {
            return step + 1;
        }
        for (auto v : mp[u])
        {
            if (!vis[v])
            {
                q.push(make_pair(v, step + 1));
                vis[v] = 1;
            }
        }
    }
    return -1;
}
int n, m;

void init(int num)
{
    for (int i = 1; i <= num; i++)
        mp[i].clear(), vis[i] = 0;
}
void solve()
{
    int r, c;
    scanf("%d %d", &r, &c);
    getchar();
    n = r * 4 + 3;
    m = 6 * c + 3;
    for (int i = 0; i <= n + 20; i++)
        for (int j = 0; j <= m + 20; j++)
            s[i][j] = 0;
    for (int i = 1; i <= n; i++)
    {
        gets(s[i] + 1);
        // cout << (s[i] + 1) << endl;
    }
    int ct = 0;
    for (int i = 3; i <= n; i += 4)
        for (int j = 5; j <= m; j += 12)
        {
            id[i][j] = ++ct;
            if (s[i][j] == 'S')
                sid = ct;
            else if (s[i][j] == 'T')
                eid = ct;
            for (int k = 0; k < 3; k++)
                id[i - mov[k][0]][j - mov[k][1]] = ct;
        }

    for (int i = 5; i <= n; i += 4)
        for (int j = 11; j <= m; j += 12)
        {
            id[i][j] = ++ct;
            if (s[i][j] == 'S')
                sid = ct;
            else if (s[i][j] == 'T')
                eid = ct;
            for (int k = 0; k < 3; k++)
                id[i - mov[k][0]][j - mov[k][1]] = ct;
        }

    for (int i = 3; i <= n; i += 4)
    {
        for (int j = 5; j <= m; j += 12)
        {
            for (int k = 0; k < 3; k++)
            {
                int xx = i + mov[k][0];
                int yy = j + mov[k][1];
                if (s[xx][yy] == ' ')
                    add(id[i][j], id[xx][yy]);
            }
        }
    }
    for (int i = 5; i <= n; i += 4)
    {
        for (int j = 11; j <= m; j += 12)
        {
            for (int k = 0; k < 3; k++)
            {
                int xx = i + mov[k][0];
                int yy = j + mov[k][1];
                if (s[xx][yy] == ' ')
                    add(id[i][j], id[xx][yy]);
            }
        }
    }
    printf("%d\\n", bfs());
    init(ct);
}
int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
        solve();

    return 0;
}