[CF590C] Three States - BFS

[CF590C] Three States - BFS

Description

给你一个?×?的地图，’.’为荒地，’#’为石头，数字为国家，求最少将多少个荒地修为路可以将三个国家联通。无解输出‘-1’。

Solution

四个点的树，四种可能

需要计算出每个点到每个连通块的最近距离，以及连通块之间相互的最近距离

#include <bits/stdc++.h>
using namespace std;

#define int long long

const int N = 1005;

#define MAP_EMPTY '.'
#define MAP_WALL '#'
#define MAP_BLOCK(x) ('0' + x)

char a[N][N];
int n, m;

int d[4][N][N], cod[4][4];

const int adi[4] = {0, 0, -1, 1};
const int adj[4] = {1, -1, 0, 0};

signed main()
{
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i] + 1;
    memset(d, 0x1f, sizeof d);
    memset(cod, 0x1f, sizeof cod);
    for (int t = 1; t <= 3; t++)
    {
        queue<pair<int, int>> que;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                if (a[i][j] == MAP_BLOCK(t))
                    que.push({i, j}), d[t][i][j] = 0;
        while (que.size())
        {
            auto [i, j] = que.front();
            que.pop();
            for (int dir = 0; dir < 4; dir++)
            {
                int di = adi[dir];
                int dj = adj[dir];
                int ni = di + i;
                int nj = dj + j;
                if (ni > 0 && nj > 0 && ni <= n && nj <= m && d[t][i][j] + 1 < d[t][ni][nj] && a[ni][nj] != MAP_WALL)
                {
                    d[t][ni][nj] = d[t][i][j] + 1;
                    que.push({ni, nj});
                }
            }
        }
    }
    for (int x = 1; x <= 3; x++)
    {
        for (int y = 1; y <= 3; y++)
        {
            if (x != y)
            {
                for (int i = 1; i <= n; i++)
                {
                    for (int j = 1; j <= m; j++)
                    {
                        if (a[i][j] == MAP_BLOCK(y))
                            cod[x][y] = min(cod[x][y], d[x][i][j]);
                    }
                }
            }
        }
    }
    int ans = 1e18;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            ans = min(ans, d[1][i][j] + d[2][i][j] + d[3][i][j] - 2);
        }
    }
    ans = min(ans, cod[1][2] + cod[2][3] - 2);
    ans = min(ans, cod[2][1] + cod[1][3] - 2);
    ans = min(ans, cod[1][3] + cod[3][2] - 2);
    if (ans < 1e12)
        cout << ans << endl;
    else
        cout << -1;
}