[POJ 3635] Full Tank? (多维最短路）

题目链接：http://poj.org/problem?id=3635

题目大意：

在一个国家有N座城市，有M条道路连接N座城市，每条道路有长度d，一单位长度耗一单位油。在每座城市有加油站，一单位价格为pi。

现在有q个询问，每个询问代表一辆车从城市st到城市ed的最少花费，其中每辆车的邮箱最大为c。

解题思路：

将每座城市拆分为c个状态，要么在这里加一单位油，要么从该点走向其他城市。用二维数组表示vis[N][C]该点是否已经访问过。

/*
ID: [email protected]
PROG: beads
LANG: C++
*/
#include<iostream>
#include<fstream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<ctype.h>
#include<algorithm>
#include<string>
#define PI acos(-1.0)
#define maxn 1010
#define INF 1<<25
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
using namespace std;
int n, m, vis[maxn][105], fuel[maxn], s, e, c;
struct node
{
    int x, d, w;
    friend bool operator < (node s, node v)
    {
        return s.w > v.w;
    }
};
vector<node> V[maxn];
int bfs()
{
    priority_queue<node> q;
    mem(vis, 0);
    node now, next;
    now.x = s, now.d = 0, now.w = 0;
    q.push(now);
    while(!q.empty())
    {
        now = q.top();
//        cout<<now.x<<" "<<now.d<<" "<<now.w<<endl;
        q.pop();
        vis[now.x][now.d] = 1;
        if (now.x == e) return now.w;
        if (!vis[now.x][now.d + 1] && now.d + 1 <= c)
        {
            next.x = now.x, next.d = now.d + 1, next.w =now.w + fuel[now.x];
            q.push(next);
        }
        for (int i = 0; i < V[now.x].size(); i++)
        {
            int u = now.x, v = V[now.x][i].x, l = V[now.x][i].d;
            if (now.d >= l && !vis[v][now.d - l])
            {
                next.x = v, next.d = now.d - l, next.w = now.w;
                q.push(next);
            }
        }
    }
    return -1;
}
int main ()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) scanf("%d", fuel + i);
    node tmp;
    int u, v;
    while(m--)
    {
        scanf("%d%d%d", &u, &v, &tmp.d);
        tmp.x = v, V[u].push_back(tmp);
        tmp.x = u, V[v].push_back(tmp);
    }
    scanf("%d", &m);
    while(m--)
    {
        scanf("%d%d%d", &c, &s, &e);
        int ans = bfs();
        if (ans == -1) puts("impossible");
        else printf("%d\n", ans);
    }
    return 0;
}