18.12.20 DSA Full Tank?（DP+BFS）

描述

After going through the receipts from your car trip through Europe this summer, you realised that the gas prices varied between the cities you visited. Maybe you could have saved some money if you were a bit more clever about where you filled your fuel?

To help other tourists (and save money yourself next time), you want to write a program for finding the cheapest way to travel between cities, filling your tank on the way. We assume that all cars use one unit of fuel per unit of distance, and start with an empty gas tank.

输入

The first line of input gives 1 ≤ n ≤ 1000 and 0 ≤ m ≤ 10000, the number of cities and roads. Then follows a line with n integers 1 ≤ p_i ≤ 100, where p_i is the fuel price in the ith city. Then follow m lines with three integers 0 ≤ u, v < n and 1 ≤ d ≤ 100, telling that there is a road between u and v with length d. Then comes a line with the number 1 ≤ q ≤ 100, giving the number of queries, and q lines with three integers 1 ≤ c ≤ 100, s and e, where c is the fuel capacity of the vehicle, s is the starting city, and e is the goal.

输出

For each query, output the price of the cheapest trip from s to e using a car with the given capacity, or "impossible" if there is no way of getting from s to e with the given car.

样例输入

5 5
10 10 20 12 13
0 1 9
0 2 8
1 2 1
1 3 11
2 3 7
2
10 0 3
20 1 4

样例输出

170
impossible

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stack>
#include <string>
#include <math.h>
#include <queue>
#include <stdio.h>
#include <string.h>
#include <set>
#include <vector>
#include <fstream>
#define maxn 1005
#define inf 999999
#define cha 127
#define eps 1e-6 
#define oo 1503
using namespace std;

int n, m;
int price[maxn];
struct node {
    int e, len;
    node(int a, int b) :e(a), len(b) {}
};
struct state {
    int city, remain, cost;
    state(int a, int b, int c) :city(a), remain(b), cost(c) {}
};
bool operator <(state a, state b) {
    return a.cost > b.cost;
}
priority_queue<state>steps;
vector<vector<node>>G(1005);
int dp[1005][105];
int c, s, e;

void f() {
    while (!steps.empty())steps.pop();
    steps.push(state(s, 0, 0));
    while (!steps.empty()) {
        state k = steps.top();
        if (k.city == e) {
            printf("%d
", k.cost);
            return;
        }
        steps.pop();
        int s = k.city, remain = k.remain, cost = k.cost;
        int size = G[s].size();
        if (remain < c&&cost+price[s]<dp[s][remain+1]) {
            steps.push(state(s, remain + 1, cost + price[s]));
            dp[s][remain + 1] =  cost + price[s];
        }
        for (int i = 0; i < size; i++) {
            int len = G[s][i].len;
            if (remain >= len && cost < dp[G[s][i].e][remain - len])
            {
                steps.push(state(G[s][i].e, remain - len, cost));
                dp[G[s][i].e][remain - len] = cost;
            }
        }
    }
    printf("impossible
");
}

void init() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++)
        scanf("%d", &price[i]);
    for (int i = 1; i <= m; i++) {
        int s, e, len;
        scanf("%d%d%d", &s, &e, &len);
        G[s].push_back(node(e, len));
        G[e].push_back(node(s, len));
    }
    int kase;
    scanf("%d", &kase);
    while (kase--) {
        scanf("%d%d%d", &c, &s, &e);
        for (int i = 0; i < n; i++)
            for (int j = 0; j <= c; j++)
                dp[i][j] = inf;
        dp[0][0] = 0;
        f();
    }
}

int main() {
    init();
    return 0;
}

思路：

记录每次BFS的状态值，用优先队列来维护（cost从小到大），直到遇到某个状态到达目标城市或队列为空（没有合法值）。

感觉也并不是很最短路。