poj 3411 Paid Roads（dfs）

Description

A network of m roads connects N cities (numbered from  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 ai to city bi:

in advance, in a city ci (which may or may not be the same as ai);
after the travel, in the city bi.
The payment is Pi in the first case and Ri in the second case.

Write a program to find a minimal-cost route from the city  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 ai, bi, ci, Pi, Ri ( ≤ i ≤ m). Adjacent values on the same line are separated by one or more spaces. All values are integers,  ≤ m, N ≤ ,  ≤ Pi , Ri ≤ , Pi ≤ Ri ( ≤ i ≤ m).

Output

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

Sample Input

Sample Output

Source

Northeastern Europe 2002, Western Subregion

 

直接dfs回溯找出最小的花费即可，以为n很小

 

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 using namespace std;
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 26
 #define inf 1e12
 int n,m;

 struct Node{
     int b,c,p,r;
 };
 vector<Node> node[N];
 int ans;
 int vis[N];
 void dfs(int u,int cost){
     vis[u]++;
     if(cost>=ans) return;
     if(u==n){
         if(cost<ans)
           ans=cost;
           return;
     }

     int size=node[u].size();
     for(int i=;i<size;i++){
         int v=node[u][i].b;
         if(vis[v]<=){
             int t=;
             if(vis[node[u][i].c] && t>node[u][i].p){
                 t=node[u][i].p;
             }
             if(t>node[u][i].r){
                 t=node[u][i].r;
             }
             dfs(v,t+cost);
             vis[v]--;
         }
     }

 }
 int main()
 {
     while(scanf("%d%d",&n,&m)==){
         for(int i=;i<N;i++) node[i].clear();
         //scanf("%d%d",&n,&m);
         for(int i=;i<m;i++){
             int a,b,c,p,r;
             scanf("%d%d%d%d%d",&a,&b,&c,&p,&r);
             Node tmp;
             //tmp.a=a;
             tmp.b=b;
             tmp.c=c;
             tmp.p=p;
             tmp.r=r;
             node[a].push_back(tmp);
         }
         memset(vis,,sizeof(vis));
         ans=;
         dfs(,);
         if(ans==) printf("impossible\n");
         else printf("%d\n",ans);
     }
     return ;
 }