POJ-2688 Cleaning Robot

Question: the shortest path for recycling all garbage

Train of Thought: First BFS processes the distance between two spam, and then DFS memory-based search

Dp [I] [state] indicates that the status after I is processed is the shortest path of state.

#include 
 
  #include 
  
   #include 
   
    #include #include 
    
     #include 
     
      using namespace std;const int MAXN = 30;const int INF = 0x3f3f3f3f;struct Point {int x,y;}point[20];struct node {int x,y,step,f;node(int _x, int _y, int _step, int _f) {x = _x, y = _y, step = _step, f = _f;}bool operator< (const node a)const {return f > a.f;}};char map[MAXN][MAXN];int n,m;int dx[4] = {1,-1,0,0};intdy[4] = {0,0,1,-1};int dis[MAXN][MAXN], vis[MAXN][MAXN];int dp[12][1<<12],sum;int check(int x, int y) {if (x > 0 && x <= n && y > 0 && y <= m && map[x][y] != 'x')return 1;return 0;}int getdiff(const Point &a, const Point &b) {return abs(a.x-b.x) + abs(a.y-b.y);}int bfs(const Point &a, const Point &b) {priority_queue
      
        q;q.push(node(a.x, a.y, 0, getdiff(a, b)));memset(vis, 0, sizeof(vis));vis[a.x][a.y] = 1;while (!q.empty()) {node cur = q.top();q.pop();if (cur.x == b.x && cur.y == b.y)return cur.step;for (int i = 0; i < 4; i++) {Point tmp;tmp.x = cur.x + dx[i];tmp.y = cur.y + dy[i];if (check(tmp.x, tmp.y) && !vis[tmp.x][tmp.y]) {vis[tmp.x][tmp.y] = 1;int f = cur.step+1+getdiff(tmp, b);q.push(node(tmp.x, tmp.y, cur.step+1, f));}}}return -1;}int getFlag(int i) {return 1<