Network Flow, question 24, question 14, island rescue, problem hierarchy

A grid chart has some points that may have some key. There may be a door between the node and the node. Some doors can pass through with specific keys, and some cannot pass through in any way. Calculate the shortest time from (1, 1) to (m, n.

Idea: layer chart + state compression. F [I] [J] [K], where I and j describe the current location, K is the current key that is compressed (because the number of keys is <= 10, so all States can be completed within the space of 1 <10 ). Then, when updating data in four directions, determine whether the data can pass through the 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,sizeof(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 << 11); ++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][_status] > 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 false;if(need_key == -1)return true;return (status >> need_key)&1;}

Network Flow, question 24, question 14, island rescue, problem hierarchy