[CODEVS1912] 汽车加油行驶问题（分层图最短路）

Posted 2020-09-22 蒟蒻zht的博客

传送门

 

吐槽：神tm网络流

 

dis[i][j][k] 表示到 (i, j) 还有 k 油的最优解

然后跑spfa，中间分一大堆情况讨论

1.当前队头还有油

　　1.目标点有加油站——直接过去

　　2.目标点每加油站——1.直接过去

　　　　　　　　　　　　2.在当前点召唤一个加油站再过去

2.没油——召唤加油站再走

 

——代码

 1 #include <queue>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #define N 101
 6 #define min(x, y) ((x) < (y) ? (x) : (y))
 7 
 8 inline int read()
 9 {
10     int x = 0, f = 1;
11     char ch = getchar();
12     for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
13     for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
14     return x * f;
15 }
16 
17 int n = read(), k = read(), a = read(), b = read(), c = read(), ans = ~(1 << 31);
18 int map[N][N], dis[N][N][N];
19 int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0}, cos[4] = {0, 0, b, b};
20 bool vis[N][N][N];
21 
22 struct node
23 {
24     int x, y, res;
25     node(int x = 0, int y = 0, int res = 0) : x(x), y(y), res(res) {}
26 };
27 
28 std::queue <node> q;
29 
30 inline void init(int x, int y, int res, int cost)
31 {
32     if(dis[x][y][res] > cost)
33     {
34         dis[x][y][res] = cost;
35         if(!vis[x][y][res])
36         {
37             vis[x][y][res] = 1;
38             q.push(node(x, y, res));
39         }
40     }
41 }
42 
43 inline void spfa()
44 {
45     node now;
46     int i, x, y, res, cost;
47     memset(dis, 127 / 3, sizeof(dis));
48     dis[1][1][k] = 0;
49     q.push(node(1, 1, k));
50     while(!q.empty())
51     {
52         now = q.front();
53         q.pop();
54         vis[now.x][now.y][now.res] = 0;
55         for(i = 0; i < 4; i++)
56         {
57             x = now.x + dx[i];
58             y = now.y + dy[i];
59             if(!x || x > n || !y || y > n) continue;
60             cost = dis[now.x][now.y][now.res] + cos[i];
61             if(now.res)
62             {
63                 if(map[x][y]) init(x, y, k, cost + a);
64                 else
65                 {
66                     init(x, y, now.res - 1, cost);
67                     init(x, y, k - 1, cost + a + c);
68                 }
69             }
70             else init(x, y, k - 1, cost + a + c);
71         }
72     }
73 }
74 
75 int main()
76 {
77     int i, j;
78     for(i = 1; i <= n; i++)
79         for(j = 1; j <= n; j++)
80             map[i][j] = read();
81     spfa();
82     for(i = 0; i <= k; i++) ans = min(ans, dis[n][n][i]);
83     printf("%d\n", ans);
84     return 0;
85 }

View Code