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); else printf("impossible/n"); } return 0; }