Paid Roads

Problem 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 5 1 2 1 10 10 2 3 1 30 50 3 4 3 80 80 2 1 2 10 10 1 3 2 10 50

 

Sample Output

110

 **************************************************************************************

dfs&&标记

***************************************************************************************

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
const  int inf=1<<28;
struct  node
{
    int u,v,c,p,r;
};
vector<struct node>g[120];
int vis[1001];
int n,m,i,j,k;
int ans;
int u,v,c,p,r;
void dfs(int x,int val)
 {
     vis[x]++;//访问次数
     if(x==n)
      {
          if(ans>val)//到底返回
           ans=val;
          return;
      }
      if(ans<val)
       return;
      for(int it=0;it<g[x].size();it++)
       {
           int v=g[x][it].v;
           int c=g[x][it].c;
           if(vis[v]<=3)//子节点访问不多于三次
            {
                int min=inf;
                if(vis[c]&&min>g[x][it].p)
                  min=g[x][it].p;
                if(min>g[x][it].r)
                  min=g[x][it].r;
                dfs(v,val+min);
                vis[v]--;//恢复
            }
       }
 }
 int main()
 {
     cin>>n>>m;
     for(i=1;i<=m;i++)
      {
          cin>>u>>v>>c>>p>>r;
          g[u].push_back((struct node){u,v,c,p,r});//存储（重点）
      }
      memset(vis,0,sizeof(vis));
      ans=inf;
      dfs(1,0);
      if(ans==inf)
       cout<<"impossible"<<endl;
       else
        cout<<ans<<endl;
      return 0;
 }