【题解】
用dijkstra算法求最短路。同时考虑在每个节点加油（一单位）与否。dp[i][j]记录汽车到达第i个点所剩j单位油所用的费用。
【代码】
1 #include <iostream> 2 #include <map> 3 #include <cstring> 4 #include <string> 5 #include <queue> 6 using namespace std; 7 #define maxn 1005 8 #define maxm 10005 9 10 /* 11 * dp[i][j] denodes the least cost on the ith node 12 * with j oil left. 13 */ 14 int price[maxn], dp[maxn][105], vis[maxn][105]; 15 int Ecnt, capacaity, sp, ep; 16 17 struct Node 18 { 19 int node, cost, oil; 20 Node(int n, int c, int o) : 21 node(n), cost(c), oil(o) {} 22 bool operator < (const Node &N) const{ 23 return cost > N.cost; 24 } 25 }; 26 27 struct Edge 28 { 29 int node; 30 int len; 31 Edge* next; 32 }edges[2*maxm]; 33 34 Edge* head[maxn]; 35 36 void init() 37 { 38 Ecnt = 0; 39 fill(head, head + maxn, (Edge*)0); 40 } 41 42 void build(int u, int v, int w) 43 { 44 edges[Ecnt].node = u; 45 edges[Ecnt].len = w; 46 edges[Ecnt].next = head[v]; 47 head[v] = edges + Ecnt++; 48 49 edges[Ecnt].node = v; 50 edges[Ecnt].len = w; 51 edges[Ecnt].next = head[u]; 52 head[u] = edges + Ecnt++; 53 } 54 55 void dijkstra() 56 { 57 memset(vis, 0, sizeof(vis)); 58 memset(dp, 1, sizeof(dp)); 59 priority_queue<Node> pq; 60 pq.push(Node(sp, 0, 0)); 61 dp[sp][0] = 0; 62 while (!pq.empty()) { 63 Node np = pq.top(); 64 pq.pop(); 65 int n = np.node, c = np.cost, o = np.oil; 66 vis[n][o] = 1; 67 if (n == ep) { 68 printf("%d ", c); 69 return; 70 } 71 /* decide to add oil or not */ 72 if (o + 1 <= capacaity && !vis[n][o + 1] && dp[n][o] + price[n] < dp[n][o + 1]) { 73 dp[n][o + 1] = dp[n][o] + price[n]; 74 pq.push(Node(n, dp[n][o + 1], o + 1)); 75 } 76 /* drive to the next node */ 77 for (Edge *Ep = head[n]; Ep; Ep = Ep->next) { 78 int N = Ep->node, Len = Ep->len; 79 if (o >= Len && !vis[N][o - Len] && c < dp[N][o - Len]) { 80 dp[N][o - Len] = c; 81 pq.push(Node(N, dp[N][o - Len], o - Len)); 82 } 83 } 84 } 85 printf("impossible "); 86 return; 87 } 88 89 int main() 90 { 91 int city_n, road_m, queries; 92 cin >> city_n >> road_m; 93 init(); 94 for (int i = 0; i < city_n; i++) 95 cin >> price[i]; 96 for (int i = 0; i < road_m; i++) { 97 int u, v, d; 98 cin >> u >> v >> d; 99 build(u, v, d); 100 } 101 cin >> queries; 102 for (int i = 0; i < queries; i++) { 103 cin >> capacaity >> sp >> ep; 104 dijkstra(); 105 } 106 //system("pause"); 107 return 0; 108 }