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.

输入描述

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 ≤ pi ≤ 100, where pi 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.

输出描述

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.

输入例子

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

输出例子

170
impossible

参考答案

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#define MAXN 1005
#define MAXM 100005
#define INF 1000000000
using namespace std;
struct Edge
{
    int v, w, next;
}edge[MAXM];
struct Node
{
    int v, cost, f;
    bool operator <(const Node &a) const{
        return a.cost < cost;
    }
};
int head[MAXN], e, n, m, cap;
int dp[MAXN][105], used[MAXN][105], p[MAXN];
int s, t, ask;
priority_queue<Node>q;
void init()
{
    memset(head, -1, sizeof(head));
    e = 0;
}
void ready()
{
    for(int i = 0; i < n; i++)
        for(int j = 0; j <= 100; j++)
            dp[i][j] = INF;
    dp[s][0] = 0;
    memset(used, 0, sizeof(used));
    while(!q.empty()) q.pop();
}
void insert(int x, int y, int w)
{
    edge[e].v = y;
    edge[e].w = w;
    edge[e].next = head[x];
    head[x] = e++;
}
int bfs()
{
    Node a, b;
    a.v = s, a.cost = 0, a.f = 0;
    q.push(a);
    while(!q.empty())
    {
        a = q.top();
        q.pop();
        int u = a.v;
        int cost = a.cost;
        int f = a.f;
        used[u][f] = 1;
        if(u == t) return cost;
        if(f + 1 <= cap && !used[u][f + 1] && dp[u][f + 1] > dp[u][f] + p[u])
        {
            dp[u][f + 1] = dp[u][f] + p[u];
            b.v = u;
            b.f = f + 1;
            b.cost = dp[u][f + 1];
            q.push(b);
        }
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].v;
            int w = edge[i].w;
            if(f >= w && !used[v][f - w] && dp[v][f - w] > cost)
            {
                dp[v][f - w] = cost;
                b.v = v;
                b.f = f - w;
                b.cost = dp[v][f - w];
                q.push(b);
            }
        }
    }
    return -1;
}
int main()
{
    int x, y, w;
    scanf("%d%d", &n, &m);
    init();
    for(int i = 0; i < n; i++)  scanf("%d", &p[i]);
    while(m--)
    {
        scanf("%d%d%d", &x, &y, &w);
        insert(x, y, w);
        insert(y, x, w);
    }
    scanf("%d", &ask);
    while(ask--)
    {
        scanf("%d%d%d", &cap, &s, &t);
        ready();
        int ans = bfs();
        if(ans != -1) printf("%d\n", ans);
        else printf("impossible\n");
    }
    return 0;
}