Luogu P4011 孤岛营救问题

题目链接 \(Click\) \(Here\)

注意坑点：一个地方可以有多把钥匙。

被卡了一会，调出来发现忘了取出来实际的数字，直接把二进制位或上去了\(TwT\)，其他的就是套路的分层图最短路。不算太难。

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

int n, m, p, k, s;
int can[11][11][11][11], key[11];
int mv[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

int node (int x, int y, int f) {
    return f * n * m + (x - 1) * m + y;
}

bool in_map (int x, int y) {
    return 1 <= x && x <= n && 1 <= y && y <= m;
}

const int N = 1000010;
const int M = 4000010;

int cnt, head[N];
const int INF = 0x3f3f3f3f;

struct edge {
    int nxt, to, w;
    edge (int _nxt = 0, int _to = 0, int _w = 0) {
        nxt = _nxt, to = _to, w = _w;
    }
}e[M];

void add_edge (int u, int v, int w) {
    e[++cnt] = edge (head[u], v, w); head[u] = cnt;
}

queue <int> q;
int vis[N], dis[N];

int spfa (int s, int t) {
    memset (dis, 0x3f, sizeof (dis));
    vis[s] = true; dis[s] = 0; q.push (s);
    while (!q.empty ()) {
        int u = q.front (); q.pop ();
        for (int i = head[u]; i; i = e[i].nxt) {
            int v = e[i].to;
            if (dis[v] > dis[u] + e[i].w) {
                dis[v] = dis[u] + e[i].w;
                if (!vis[v]) {
                    vis[v] = true;
                    q.push (v);
                }
            }
        }
        vis[u] = false;
    }
    return dis[t] == INF ? -1 : dis[t];
}

vector <int> have[11][11];

int main () {
    cin >> n >> m >> p >> k;
    for (int i = 1; i <= k; ++i) {
        static int x1, x2, y1, y2, g;
        cin >> x1 >> y1 >> x2 >> y2 >> g;
        if (g == 0) {
            can[x1][y1][x2][y2] = -1;
            can[x2][y2][x1][y1] = -1;
        } else {
            can[x1][y1][x2][y2] = g;
            can[x2][y2][x1][y1] = g;
        }
    }
    cin >> s;
    for (int i = 1; i <= s; ++i) {
        static int x, y, g;
        cin >> x >> y >> g;
        have[x][y].push_back (g);
    }
    int s = node (n, m, (1 << p)) + 1;
    int t = node (n, m, (1 << p)) + 2;
    for (int i = 0; i < (1 << p); ++i) {
        memset (key, 0, sizeof (key));
        for (int t = i, wei = 1; t != 0; t >>= 1, ++wei) {
            key[wei] = t & 1;
        } 
        for (int x = 1; x <= n; ++x) {
            for (int y = 1; y <= m; ++y) {
                for (int t = 0; t < 4; ++t) {
                    int tx = x + mv[t][0];
                    int ty = y + mv[t][1];
                    if (!in_map (tx, ty)) continue;
                    if (can[x][y][tx][ty] == 0 || (can[x][y][tx][ty] > 0 && key[can[x][y][tx][ty]] != 0)) {
                        // 没有门 /　有钥匙
                        // if (can[x][y][tx][ty] != 0) printf ("node (%d, %d, %d) -> node (%d, %d, %d)\n", x, y, i, tx, ty, i);
                        add_edge (node (x, y, i), node (tx, ty, i), 1);
                    }
                }
                for (int t = 0; t < have[x][y].size (); ++t) {
                    if (!key[have[x][y][t]])
                        add_edge (node (x, y, i), node (x, y, (i | (1 << (have[x][y][t] - 1)))), 0);
                }
            }
        }
        add_edge (node (n, m, i), t, 0);
    }
    add_edge (s, node (1, 1, 0), 0);
    cout << spfa (s, t) << endl;
}