ACM: 深搜题 poj 3411

                                                               
Paid Roads

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
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 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 ai,
bi, ci,
Pi, Ri (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 ≤
Pi , Ri ≤
100, Pi ≤
Ri (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

题意: 现在给你n个城市m条路的图. 部分路要收费的. 收费方式有2种: 经过Ci城市的就Pi , 否则Ri.

        
现在要你求出最小的费用从1 到 n城市.

 

解题思路:

                 
1. 条件最短路. 标记神搜即可了.

                 
2. 每个点经过不超过三次.(关键,使深搜树有了最下层)

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

using namespace std;

#define MAX 15

const int INF = 1 << 30;

struct node

{

    int v;

    int c;

    int p;

    int r;

    int
next;

}edges[MAX];

int n, m;

int next[MAX];

int vis[MAX];

int result;

int num;

void init()

{

    num =
0;

    result =
INF;

  
 memset(next,-1,sizeof(next));

  
 memset(vis,0,sizeof(vis));

}

void dfs(int v,int w)

{

    if(w
> INF)

  
   
 return ;

    if(v ==
n)

    {

  
   
 if(result > w)

  
   
   
 result = w;

  
   
 return ;

    }

    

    for(int e =
next[v]; e != -1; e = edges[e].next)

    {

  
   
 int c = edges[e].c;

  
   
 int p = edges[e].p;

  
   
 int r = edges[e].r;

  
   
 

  
   
 if(vis[ edges[e].v ] <= 2)

  
   
 {

  
   
   
 vis[ edges[e].v ]++;

  
   
   
 if(vis[c] > 0)

  
   
   
   
 dfs(edges[e].v,w+p);

  
   
   
 else

  
   
   
   
 dfs(edges[e].v,w+r);

  
   
   
 vis[ edges[e].v ]--;

  
   
 }

  
   
 

    }

}

int main()

{

//  
 freopen("input.txt","r",stdin);

    int a, b, c,
p, r;

  
 while(scanf("%d
%d",&n,&m) != EOF)

    {

  
   
 init();

  
   
 for(int i = 1; i <= m; ++i)

  
   
 {

  
   
   
 scanf("%d %d %d %d
%d",&a,&b,&c,&p,&r);

  
   
   
 edges[num].v = b;

  
   
   
 edges[num].c = c;

  
   
   
 edges[num].p = p;