UVa 1600 Patrol Robot (习题 6-5)

Posted 2020-07-02

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

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef struct Node{
 5     int x, y, step, k;
 6 } node;
 7 const int MAXN = 25;
 8 const int INF = 0x3f3f3f3f;
 9 const int cx[] = {-1, 1, 0, 0};
10 const int cy[] = {0, 0, -1, 1};
11 int n, m, k, ans;
12 int maze[MAXN][MAXN];
13 int vis[MAXN][MAXN][MAXN];
14 
15 void bfs(){
16     queue<node> q;
17     while(!q.empty()) q.pop();
18     node cur ;
19     cur.x = cur.y = 1;
20     cur.step = 0;
21     cur.k = k;
22     q.push(cur);
23     vis[cur.x][cur.y][cur.k] = 1;
24     while(!q.empty()){
25         cur = q.front();
26         q.pop();
27         int x = cur.x, y = cur.y;
28         if(x == n && y == m){
29             ans = cur.step;
30             return ;
31         }
32         node now;
33         if(cur.k >= 0){
34             for(int i = 0; i < 4; ++i){
35                 now.x = x + cx[i], now.y = y + cy[i];
36                 now.step = cur.step + 1;
37                 now.k = maze[now.x][now.y] ? cur.k - 1 : k;
38                 if(now.x < 1 || n < now.x || now.y < 1 || m < now.y || vis[now.x][now.y][now.k]) continue;
39                 if(now.k >= 0){
40                     vis[now.x][now.y][now.k] = 1;
41                     q.push(now);
42                 }
43             }
44         }
45     }
46 }
47 
48 int main(){
49     int t;
50     cin >> t;
51     while(t--){
52         cin >> n >> m >> k;
53         memset(vis, 0, sizeof(vis));
54         memset(maze, 0, sizeof(maze));
55         for(int i = 1; i <= n; ++i)
56             for(int j = 1; j <= m; ++j)
57                 cin >> maze[i][j];
58         ans = -1;
59         bfs();
60         cout << ans << endl;
61     }
62     return 0;
63 }