POJ3635 Full Tank?

【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】

 #include <iostream>
 #include <map>
 #include <cstring>
 #include <string>
 #include <queue>
 using namespace std;
 #define maxn 1005
 #define maxm 10005

 /*
 * dp[i][j] denodes the least cost on the ith node
 * with j oil left.
 */
 int price[maxn], dp[maxn][], vis[maxn][];
 int Ecnt, capacaity, sp, ep;

 struct Node
 {
     int node, cost, oil;
     Node(int n, int c, int o) :
         node(n), cost(c), oil(o) {}
     bool operator < (const Node &N) const{
         return cost > N.cost;
     }
 };

 struct Edge
 {
     int node;
     int len;
     Edge* next;
 }edges[*maxm];

 Edge* head[maxn];

 void init()
 {
     Ecnt = ;
     fill(head, head + maxn, (Edge*));
 }

 void build(int u, int v, int w)
 {
     edges[Ecnt].node = u;
     edges[Ecnt].len = w;
     edges[Ecnt].next = head[v];
     head[v] = edges + Ecnt++;

     edges[Ecnt].node = v;
     edges[Ecnt].len = w;
     edges[Ecnt].next = head[u];
     head[u] = edges + Ecnt++;
 }

 void dijkstra()
 {
     memset(vis, , sizeof(vis));
     memset(dp, , sizeof(dp));
     priority_queue<Node> pq;
     pq.push(Node(sp, , ));
     dp[sp][] = ;
     while (!pq.empty()) {
         Node np = pq.top();
         pq.pop();
         int n = np.node, c = np.cost, o = np.oil;
         vis[n][o] = ;
         if (n == ep) {
             printf("%d\n", c);
             return;
         }
         /* decide to add oil or not */
         if (o +  <= capacaity && !vis[n][o + ] && dp[n][o] + price[n] < dp[n][o + ]) {
             dp[n][o + ] = dp[n][o] + price[n];
             pq.push(Node(n, dp[n][o + ], o + ));
         }
         /* drive to the next node */
         for (Edge *Ep = head[n]; Ep; Ep = Ep->next) {
             int N = Ep->node, Len = Ep->len;
             if (o >= Len && !vis[N][o - Len] && c < dp[N][o - Len]) {
                 dp[N][o - Len] = c;
                 pq.push(Node(N, dp[N][o - Len], o - Len));
             }
         }
     }
     printf("impossible\n");
     return;
 }

 int main()
 {
     int city_n, road_m, queries;
     cin >> city_n >> road_m;
     init();
     for (int i = ; i < city_n; i++)
         cin >> price[i];
     for (int i = ; i < road_m; i++) {
         int u, v, d;
         cin >> u >> v >> d;
         build(u, v, d);
     }
     cin >> queries;
     for (int i = ; i < queries; i++) {
         cin >> capacaity >> sp >> ep;
         dijkstra();
     }
     //system("pause");
     return ;
 }