HDU1429 - 胜利大逃亡（续）

Posted 2020-08-20

与上一道题：HDU1885 - Key Task基本没什么两样

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <string>
 6 #include <queue>
 7 #include <cctype>
 8 using namespace std;
 9 const int maxn = 25;
10 int n, m, fail;
11 char G[maxn][maxn];
12 int vis[maxn][maxn][2000];
13 const int dr[] = {-1, 0, 1, 0};
14 const int dc[] = { 0, 1, 0,-1};
15 
16 struct Node
17 {
18     int r, c, key, step;
19     Node(int r = 0, int c = 0, int key = 0, int step = 0):r(r), c(c), key(key), step(step){}
20 };
21 
22 bool isInside(int r, int c)
23 {
24     if(r > 0 && r <= n && c > 0 && c <= m && G[r][c] != ‘*‘)
25         return true;
26     else
27         return false;
28 }
29 
30 void bfs(int rr, int cc)
31 {
32     queue<Node> q;
33     memset(vis, 0, sizeof vis);
34     Node node(rr, cc, 0, 0);
35     q.push(node);
36     while(!q.empty()) {
37         Node u = q.front();
38         q.pop();
39         if(G[u.r][u.c] == ‘^‘) {
40             if(u.step < fail)
41                 printf("%d\n", u.step);
42             else
43                 printf("-1\n");
44             return;
45         }
46         for(int i = 0; i < 4; i++) {
47             Node v(u.r + dr[i], u.c + dc[i], u.key, u.step+1);
48             if(isInside(v.r, v.c)) {
49                 if(islower(G[v.r][v.c])) {
50                     int k = G[v.r][v.c] - ‘a‘;
51                     if((v.key & (1 << k)) == 0) {
52                         v.key += (1 << k);
53                     }
54                     if(!vis[v.r][v.c][v.key]) {
55                         vis[v.r][v.c][v.key] = 1;
56                         q.push(v);
57                     }
58                 }
59                 else if(isupper(G[v.r][v.c])) {
60                     int k = G[v.r][v.c] - ‘A‘;
61                     if(v.key & (1 << k) && !vis[v.r][v.c][v.key]) {
62                         vis[v.r][v.c][v.key] = 1;
63                         q.push(v);
64                     }
65                 }
66                 else {
67                     if(!vis[v.r][v.c][v.key]) {
68                         vis[v.r][v.c][v.key] = 1;
69                         q.push(v);
70                     }
71                 }
72             }
73         }
74     }
75     printf("-1\n");
76 }
77 int main()
78 {
79    // freopen("in.txt", "r", stdin);
80     while(scanf("%d %d %d", &n, &m, &fail) != EOF) {
81         memset(G, ‘*‘, sizeof G);
82         int x, y;
83         for(int i = 1; i <= n; i++) {
84             scanf("%s", G[i]+1);
85             G[i][m+1] = ‘*‘;
86             G[i][m+2] = ‘\0‘;
87             if(strchr(G[i], ‘@‘)) {
88                 x = i;
89                 y = strchr(G[i], ‘@‘) - *G - i*25;
90             }
91         }
92         //printf("%d %d\n", x, y);
93         bfs(x, y);
94     }
95     return 0;
96 }