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

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

 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<cstdio>
 7 #include<queue>
 8 #include<vector>
 9 #include<stack>
10 using namespace std;
11 const  int inf=1<<28;
12 struct  node
13 {
14     int u,v,c,p,r;
15 };
16 vector<struct node>g[120];
17 int vis[1001];
18 int n,m,i,j,k;
19 int ans;
20 int u,v,c,p,r;
21 void dfs(int x,int val)
22  {
23      vis[x]++;//访问次数
24      if(x==n)
25       {
26           if(ans>val)//到底返回
27            ans=val;
28           return;
29       }
30       if(ans<val)
31        return;
32       for(int it=0;it<g[x].size();it++)
33        {
34            int v=g[x][it].v;
35            int c=g[x][it].c;
36            if(vis[v]<=3)//子节点访问不多于三次
37             {
38                 int min=inf;
39                 if(vis[c]&&min>g[x][it].p)
40                   min=g[x][it].p;
41                 if(min>g[x][it].r)
42                   min=g[x][it].r;
43                 dfs(v,val+min);
44                 vis[v]--;//恢复
45             }
46        }
47  }
48  int main()
49  {
50      cin>>n>>m;
51      for(i=1;i<=m;i++)
52       {
53           cin>>u>>v>>c>>p>>r;
54           g[u].push_back((struct node){u,v,c,p,r});//存储（重点）
55       }
56       memset(vis,0,sizeof(vis));
57       ans=inf;
58       dfs(1,0);
59       if(ans==inf)
60        cout<<"impossible"<<endl;
61        else
62         cout<<ans<<endl;
63       return 0;
64  }

View Code

转载于:https://www.cnblogs.com/sdau--codeants/p/3273416.html

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