[CODEVS1912] 汽车加油行驶问题（分层图最短路）

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

 #include <queue>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #define N 101
 #define min(x, y) ((x) < (y) ? (x) : (y))

 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;
 }

 int n = read(), k = read(), a = read(), b = read(), c = read(), ans = ~( << );
 int map[N][N], dis[N][N][N];
 int dx[] = {, , , -}, dy[] = {, , -, }, cos[] = {, , b, b};
 bool vis[N][N][N];

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

 std::queue <node> q;

 inline void init(int x, int y, int res, int cost)
 {
     if(dis[x][y][res] > cost)
     {
         dis[x][y][res] = cost;
         if(!vis[x][y][res])
         {
             vis[x][y][res] = ;
             q.push(node(x, y, res));
         }
     }
 }

 inline void spfa()
 {
     node now;
     int i, x, y, res, cost;
     memset(dis,  / , sizeof(dis));
     dis[][][k] = ;
     q.push(node(, , k));
     while(!q.empty())
     {
         now = q.front();
         q.pop();
         vis[now.x][now.y][now.res] = ;
         for(i = ; i < ; i++)
         {
             x = now.x + dx[i];
             y = now.y + dy[i];
             if(!x || x > n || !y || y > n) continue;
             cost = dis[now.x][now.y][now.res] + cos[i];
             if(now.res)
             {
                 if(map[x][y]) init(x, y, k, cost + a);
                 else
                 {
                     init(x, y, now.res - , cost);
                     init(x, y, k - , cost + a + c);
                 }
             }
             else init(x, y, k - , cost + a + c);
         }
     }
 }

 int main()
 {
     int i, j;
     for(i = ; i <= n; i++)
         for(j = ; j <= n; j++)
             map[i][j] = read();
     spfa();
     for(i = ; i <= k; i++) ans = min(ans, dis[n][n][i]);
     printf("%d\n", ans);
     return ;
 }