[CODEVS1911] 孤岛营救问题（分层图最短路）

传送门

吐槽：神tm网络流。。。

用持有的钥匙分层，状态压缩，用 2 进制表示持有的钥匙集合。

dis[i][j][k] 表示持有的钥匙集合为 k，到达点 (i, j) 的最短路径。

分层图的最短路听上去很玄乎，其实通过代码来看还是很好理解的。

——代码

 #include <queue>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #define N 20
 #define min(x, y) ((x) < (y) ? (x) : (y))

 int n, m, p, ans = ~( << );
 int map[N][N][N][N], key[N][N], dis[N][N][ << ];
 int dx[] = {, , , -}, dy[] = {, , -, };
 bool vis[N][N][ << ];

 struct node
 {
     int x, y, s;
     node(int x = , int y = , int s = ) : x(x), y(y), s(s) {}
 };

 std::queue <node> q;

 inline int read()
 {
     int x = , f = ;
     char ch = getchar();
     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -;
     for(; isdigit(ch); ch = getchar()) x = (x << ) + (x << ) + ch - '';
     return x * f;
 }

 inline bool Acc(int x1, int y1, int x2, int y2, int s)
 {
     int need_key = map[x1][y1][x2][y2];
     if(!need_key) return ;
     if(need_key == -) return ;
     return (s >> need_key - ) & ;
 }

 inline void spfa()
 {
     node now;
     int i, s, x, y;
     memset(dis,  / , sizeof(dis));
     dis[][][] = ;
     q.push(node(, , ));
     while(!q.empty())
     {
         now = q.front();
         q.pop();
         vis[now.x][now.y][now.s] = ;
         for(i = ; i < ; i++)
         {
             x = now.x + dx[i];
             y = now.y + dy[i];
             s = now.s | key[x][y];
             if(!x || x > n || !y || y > m) continue;
             if(Acc(now.x, now.y, x, y, now.s))
                 if(dis[x][y][s] > dis[now.x][now.y][now.s] + )
                 {
                     dis[x][y][s] = dis[now.x][now.y][now.s] + ;
                     if(!vis[x][y][s])
                     {
                         vis[x][y][s] = ;
                         q.push(node(x, y, s));
                     }
                 }
         }
     }
 }

 int main()
 {
     int i, j, a, b, c, d, x, y, z, doors, keys;
     n = read();
     m = read();
     p = read();
     doors = read();
     memset(map, -, sizeof(map));
     for(i = ; i <= doors; i++)
     {
         a = read();
         b = read();
         c = read();
         d = read();
         map[a][b][c][d] = map[c][d][a][b] = read();
     }
     keys = read();
     for(i = ; i <= keys; i++)
     {
         x = read();
         y = read();
         z = read();
         key[x][y] |=  << z - ;
     }
     spfa();
     for(i = ; i < ( << ); i++) ans = min(ans, dis[n][m][i]);
     if(ans == ) ans = -;
     printf("%d\n", ans);
     return ;
 }