UVa 1600 Patrol Robot (习题 6-5)

传送门: https://uva.onlinejudge.org/external/16/1600.pdf

多状态广搜

网上题解: 给vis数组再加一维状态，表示当前还剩下的能够穿越的墙的次数，每次碰到墙，当前的k减去1,碰到0，当前的k变成最初的k。

vis[x][y][z]  x, y代表坐标  z表示k  当为真时说明该点剩余穿墙次数为k的状态已出现过

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

 typedef struct Node{
     int x, y, step, k;
 } node;
 const int MAXN = ;
 const int INF = 0x3f3f3f3f;
 const int cx[] = {-, , , };
 const int cy[] = {, , -, };
 int n, m, k, ans;
 int maze[MAXN][MAXN];
 int vis[MAXN][MAXN][MAXN];

 void bfs(){
     queue<node> q;
     while(!q.empty()) q.pop();
     node cur ;
     cur.x = cur.y = ;
     cur.step = ;
     cur.k = k;
     q.push(cur);
     vis[cur.x][cur.y][cur.k] = ;
     while(!q.empty()){
         cur = q.front();
         q.pop();
         int x = cur.x, y = cur.y;
         if(x == n && y == m){
             ans = cur.step;
             return ;
         }
         node now;
         if(cur.k >= ){
             for(int i = ; i < ; ++i){
                 now.x = x + cx[i], now.y = y + cy[i];
                 now.step = cur.step + ;
                 now.k = maze[now.x][now.y] ? cur.k -  : k;
                 if(now.x <  || n < now.x || now.y <  || m < now.y || vis[now.x][now.y][now.k]) continue;
                 if(now.k >= ){
                     vis[now.x][now.y][now.k] = ;
                     q.push(now);
                 }
             }
         }
     }
 }

 int main(){
     int t;
     cin >> t;
     while(t--){
         cin >> n >> m >> k;
         memset(vis, , sizeof(vis));
         memset(maze, , sizeof(maze));
         for(int i = ; i <= n; ++i)
             for(int j = ; j <= m; ++j)
                 cin >> maze[i][j];
         ans = -;
         bfs();
         cout << ans << endl;
     }
     return ;
 }