Network Flow 24 Questions 14 Island rescue problem Layer Map

The main topic: a grid chart, some of the points above may have some kind of key. There may be gates between nodes and nodes. Some doors require specific keys to pass, and some do not. The shortest time to start from (m,n).

Idea: Hierarchical graph + state compression.

F[i][j][k], where I and J describe the current location. K is the current key that has been compressed (because the number of keys is <=10, so the whole state can be 1<<10 within the space). Then, when updating in four directions, it is possible to pass through a door.

CODE:

#include <queue> #include <cstdio> #include <cstring> #include <iostream> #include < algorithm> #define MAX 20#define INF 0x3f3f3f3fusing namespace Std;const int dx[] = {0,1,-1,0,0};const int dy[] = {0,0,0 , 1,-1};struct Complex{int x,y,status; Complex (int _,int __,int ___): X (_), Y (__), status (___) {}complex () {}};int m,n,p,doors,keys;int f[max][max][1 << 11 ];bool v[max][max][1 << 11];int map[max][max][max][max];int key[max][max];void SPFA (); inline bool Accelerator ( int X1,int y1,int x2,int y2,int status); int main () {cin >> m >> n >> P >> doors;memset (Map,-1,sizeo F (map)); for (int a,b,c,d,x,i = 1;i <= doors; ++i) {scanf ("%d%d%d%d%d", &a,&b,&c,&d,&x); Map[a][b] C [d] = map[c][d][a][b] = x;} CIN >> keys;for (int x,y,z,i = 1;i <= keys; ++i) {scanf ("%d%d%d", &x,&y,&z); Key[x][y] |= 1 << z;} SPFA (); int ans = inf;for (int i = 0;i < (1 << one); ++i) ans = min (ans,f[m][n][i]); if (ans = = INF) ans = -1;cout << ans << endl;return 0;} void Spfa () {memset (f,0x3f,sizeof (f)); F[1][1][0] = 0;static queue<complex> q;while (!q.empty ()) Q.pop (); Q.push ( Complex (1,1,0)); while (!q.empty ()) {Complex now = Q.front (); Q.pop (); V[now.x][now.y][now.status] = False;int _status = Now.status;if (Key[now.x][now.y]) _status |= key[now.x][now.y];for (int i = 1;i <= 4; ++i) {int FX = now.x + dx[i];int fy = Now.y + dy[i];if (!fx | |!fy | | FX > M | | fy > N) continue;if (Accelerator (Now.x,now.y,fx,fy,_status)) if (f[fx][fy][_s Tatus] > F[now.x][now.y][now.status] + 1) {F[fx][fy][_status] = F[now.x][now.y][now.status] + 1;if (!v[fx][fy][_ Status]) V[fx][fy][_status] = True,q.push (Complex (Fx,fy,_status));}}} inline bool Accelerator (int x1,int y1,int x2,int y2,int status) {int need_key = map[x1][y1][x2][y2];if (!need_key) return FA Lse;if (Need_key = =-1) return True;return (status >> need_key) &1;}

Network Flow 24 Questions 14 Island rescue problem Layer Map