1 #include <cstdio>
 2 #include <vector>
 3 #include <algorithm>
 4 #include <queue>
 5 #include <cstring>
 6 #define CLR(a, b) memset((a), (b), sizeof((a)))
 7 using namespace std;
 8 const int N = 110;
 9 const int states = 1<<11;
10 const int INF = 0x3f3f3f3f;
11
12 int d[N][states], vis[N][states];
13 int id[N][N];
14 int key[N];     //key[i]:i点有哪些钥匙（状态）
15 int mp[N][N];   //mp[i][j]:i到j有哪类门
16 int dx[] = {1, 0, -1, 0};
17 int dy[] = {0, 1, 0, -1};
18 int n, m, p;
19 struct qnode{
20     int v;
21     int x;//状态
22     qnode(int _v=0,int _x=0):v(_v),x(_x){}
23 };
24 bool SPFA(int st, int ed) {
25     CLR(vis, 0);
26     CLR(d, INF);
27     d[st][0] = 0;
28     queue<qnode> q;
29     while(!q.empty()) q.pop();
30     q.push(qnode(st, 0));
31     while(!q.empty()) {
32         qnode t = q.front(); q.pop();
33         int u = t.v;
34         int sta = t.x;
35         vis[u][sta] = false;
36
37         if(key[u]) sta |= key[u];
38         for(int i = 0; i < 4; i++) {
39             int x = (u+m-1)/m + dx[i];
40             int y = (u%m?u%m:m) + dy[i];
41
42             if(x < 1 || x > n || y < 1 || y > m) continue;
43             int v = id[x][y];
44             //不是墙 并且 有对应钥匙 或者没有门。
45             if(mp[u][v]!=0 && ( (sta&(1<<mp[u][v])) || mp[u][v]==-1)) {
46                 if(d[v][sta] > d[u][t.x] + 1) {
47                     d[v][sta] = d[u][t.x] + 1;
48                     if(!vis[v][sta]) {
49                         vis[v][sta] = true;
50                         q.push(qnode(v, sta));
51                     }
52                 }
53             }
54         }
55     }
56     int ans = INF;
57     for(int i = 0; i < states; ++i) ans = min(ans, d[ed][i]);
58     if(ans == INF) puts("-1");
59     else printf("%d\n", ans);
60 }
61 int main() {
62     int i, j, k, x1, y1, x2, y2, q, sum;
63     scanf("%d%d%d", &n, &m, &p);//行,列,门和墙的总数
64
65     int cnt = 0;
66     for(i = 1; i <= n; ++i)
67         for(j = 1; j <= m; ++j) id[i][j] = ++cnt;
68
69     CLR(mp, -1);
70     scanf("%d", &k);//门和墙总数
71     for(i = 1; i <= k; ++i) {
72         scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &q);
73         int id1 = id[x1][y1];
74         int id2 = id[x2][y2];
75         mp[id1][id2] = mp[id2][id1] = q;
76     }
77     scanf("%d", &sum);//钥匙总数
78     for(i = 1; i <= sum; ++i) {
79         scanf("%d%d%d", &x1, &y1, &q);
80         key[id[x1][y1]] |= (1<<q);
81     }
82     SPFA(1, n*m);
83     return 0;
84 }