[CODEVS1911] 孤岛营救问题（分层图最短路）

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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

 1 #include <queue>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #define N 20
 6 #define min(x, y) ((x) < (y) ? (x) : (y))
 7 
 8 int n, m, p, ans = ~(1 << 31);
 9 int map[N][N][N][N], key[N][N], dis[N][N][1 << 11];
10 int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
11 bool vis[N][N][1 << 11];
12 
13 struct node
14 {
15     int x, y, s;
16     node(int x = 0, int y = 0, int s = 0) : x(x), y(y), s(s) {}
17 };
18 
19 std::queue <node> q;
20 
21 inline int read()
22 {
23     int x = 0, f = 1;
24     char ch = getchar();
25     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
26     for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
27     return x * f;
28 }
29 
30 inline bool Acc(int x1, int y1, int x2, int y2, int s)
31 {
32     int need_key = map[x1][y1][x2][y2];
33     if(!need_key) return 0;
34     if(need_key == -1) return 1;
35     return (s >> need_key - 1) & 1;
36 }
37 
38 inline void spfa()
39 {
40     node now;
41     int i, s, x, y;
42     memset(dis, 127 / 3, sizeof(dis));
43     dis[1][1][0] = 0;
44     q.push(node(1, 1, 0));
45     while(!q.empty())
46     {
47         now = q.front();
48         q.pop();
49         vis[now.x][now.y][now.s] = 0;
50         for(i = 0; i < 4; i++)
51         {
52             x = now.x + dx[i];
53             y = now.y + dy[i];
54             s = now.s | key[x][y];
55             if(!x || x > n || !y || y > m) continue;
56             if(Acc(now.x, now.y, x, y, now.s))
57                 if(dis[x][y][s] > dis[now.x][now.y][now.s] + 1)
58                 {
59                     dis[x][y][s] = dis[now.x][now.y][now.s] + 1;
60                     if(!vis[x][y][s])
61                     {
62                         vis[x][y][s] = 1;
63                         q.push(node(x, y, s));
64                     }
65                 }    
66         }
67     }
68 }
69 
70 int main()
71 {
72     int i, j, a, b, c, d, x, y, z, doors, keys;
73     n = read();
74     m = read();
75     p = read();
76     doors = read();
77     memset(map, -1, sizeof(map));
78     for(i = 1; i <= doors; i++)
79     {
80         a = read();
81         b = read();
82         c = read();
83         d = read();
84         map[a][b][c][d] = map[c][d][a][b] = read();
85     }
86     keys = read();
87     for(i = 1; i <= keys; i++)
88     {
89         x = read();
90         y = read();
91         z = read();
92         key[x][y] |= 1 << z - 1;
93     }
94     spfa();
95     for(i = 0; i < (1 << 11); i++) ans = min(ans, dis[n][m][i]);
96     if(ans == 707406378) ans = -1;
97     printf("%d\n", ans);
98     return 0;
99 }
View Code

 

posted @ 2017-06-13 15:21  zht467  阅读(182)  评论(0)  编辑  收藏  举报