pku 3635 Full Tank? 优先队列+BFS

用二维数组state[C][fuel], 表示在城市C油量为fuel时的最小代价. 

怎样在一个节点上进行状态转移?
 1. 把该节点的油量加1, 以供继续拓展.
 2. 直接到其他可达节点. 

注意:

 1. 应使用优先队列(最小堆)优化搜索速度.

 2. 用邻接表表示一个节点所有相邻的节点, 效率更高.

 

#include <iostream>#include <cstring>#include <cstdio>#include <queue>using namespace std;struct Edge{int to, dis; //to:　尾端节点, dis: 到尾端节点的距离 Edge* next;};Edge* adj[1001]; //邻接表的N个首节点Edge edge[20001]; //所有的边（无向）int state[1001][101]; //state[C][fuel]: 在城市C油量为fuel时的最小价钱int price[1001]; //price[C]: 城市C的油价struct Node{int city, fuel, cost;Node(int _city, int _fuel, int _cost){city = _city; fuel = _fuel; cost = _cost;}bool operator<(const Node& r) const{return cost > r.cost;}};int n, m, q;int max_value, start, end;int bfs(){memset(state, 0x7f, sizeof(state));priority_queue<Node> q;q.push(Node(start, 0, 0));state[start][0] = 0;while(!q.empty()){Node n = q.top();if(n.city == end) //到达终点break;if(n.cost != state[n.city][n.fuel])continue;q.pop();//1. 把该节点的油量加1, 以供继续拓展(此节点push进queue后, 当pop出此节点可以再把油价加1...).//并更新最小代价值.if(n.fuel < max_value && state[n.city][n.fuel] + price[n.city] < state[n.city][n.fuel+1]){state[n.city][n.fuel+1] = state[n.city][n.fuel] + price[n.city];q.push(Node(n.city, n.fuel+1, state[n.city][n.fuel+1]));}//2. 直接到其他可达节点.//并更新最小代价值.for(Edge* ptr = adj[n.city]; ptr; ptr = ptr->next){if(n.fuel >= ptr->dis && state[n.city][n.fuel] < state[ptr->to][n.fuel-ptr->dis]){state[ptr->to][n.fuel-ptr->dis] = state[n.city][n.fuel];q.push(Node(ptr->to, n.fuel-ptr->dis, state[ptr->to][n.fuel-ptr->dis]));}}}return state[end][0];}int cnt;void add_edge(int u, int v, int w){Edge* ptr = &edge[cnt++];ptr->to = v;ptr->dis = w;ptr->next = adj[u];adj[u] = ptr;}int main(){scanf("%d%d", &n, &m);for(int i = 0; i < n; ++i)scanf("%d", &price[i]);int u, v, w;cnt = 0;memset(adj, 0, sizeof(adj));for(int i = 0; i < m; ++i){scanf("%d%d%d", &u, &v, &w);add_edge(u, v, w);add_edge(v, u, w);}scanf("%d", &q);for(int i = 0; i < q; ++i){scanf("%d%d%d", &max_value, &start, &end);int ret = bfs();if(ret < 0x7f7f7f7f)printf("%d/n", ret);elseprintf("impossible/n");}return 0;}