[CODEVS1912] 汽车加油行驶问题（分层图最短路）
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 }
