Paid Roads

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5481		Accepted: 1947

Description

A network of m roads connects N cities (numbered from 1 to N). There may be more than one road connecting one city with another. Some of the roads are paid. There are two ways to pay for travel on a paid road i from city a_i to city b_i:

• in advance, in a city c_i (which may or may not be the same as a_i);
• after the travel, in the city b_i.

The payment is P_i in the first case and R_i in the second case.

Write a program to find a minimal-cost route from the city 1 to the city N.

Input

The first line of the input contains the values of N and m. Each of the following m lines describes one road by specifying the values of a_i, b_i, c_i, P_i, R_i (1 ≤ i ≤ m). Adjacent values on the same line are separated by one or more spaces. All values are integers, 1 ≤ m, N ≤ 10, 0 ≤ P_i , R_i ≤ 100, P_i ≤ R_i (1 ≤ i ≤ m).

Output

The first and only line of the file must contain the minimal possible cost of a trip from the city 1 to the city N. If the trip is not possible for any reason, the line must contain the word ‘impossible’.

Sample Input

4 51 2 1 10 102 3 1 30 503 4 3 80 802 1 2 10 101 3 2 10 50

Sample Output

110

有n个城市m条路线。每条路线有5个值，a,b,c,p,r表示从a城到b城。假设到过c城，收费p元，否则收费r元。

如今从1到n，求最小花费。

这题主要是去过的点又回来的问题，比如例子1->2->1->3->4，从2又返回了1，因为m=10，所以一个点最多訪问3次(事实上取2也能A，感觉还是3靠谱点）

ps：刚開始记录了map[i][j]表示i到j能够走第map[i][j]条路。由于从第i到j能够有多条路，一直wa，再改就麻烦了，换成直接枚举m条路，反正m也不大。

代码：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;const int inf=99999999;//int map[15][15];int vis[15];int n,m;int ans;struct node{    int a;    int b;    int c;    int p;    int r;}road[15];void dfs(int u,int v){    if(v>ans)    return;    if(u==n)    {        if(v