POJ3411 DFS最小计算消费值

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

</pre><pre name="code" class="cpp">#include <IOSTREAM>
#include <CSTDLIB>
using namespace std;

#define MAX 11
#define MAXCOUNT 2000

typedef struct
{
int ai, bi, ci, Pi, Ri;
}CITY;

CITY city[MAX];
int visit[MAX];
int N, m;
int min_cost;

void search_dfs(int in_city, int cost_count)
{
if (in_city == N&&min_cost > cost_count)
{
min_cost = cost_count;
return;
}
int ic;
for (ic = 1; ic <= m; ic++)             //////////
{
if (in_city == city[ic].ai&&visit[city[ic].bi] <= 3)
{
visit[city[ic].bi]++;
if (visit[city[ic].ci] > 0) //如果曾经到过城市C则用Pi的花费计算，否则用Ri的花费计算
search_dfs(city[ic].bi, cost_count + city[ic].Pi);
else
search_dfs(city[ic].bi, cost_count + city[ic].Ri);
visit[city[ic].bi]--;
}
}
return;
}

int main()
{
while (cin >> N >> m)
{
if ((!(N >= 1 && N <= 10)) || (!(m >= 1 && m <= 10)))
continue;
memset(city, 0, sizeof(city));
memset(visit, 0, sizeof(visit));
min_cost = MAXCOUNT;
int ic;
for (ic = 1; ic <= m; ic++)
{
cin >> city[ic].ai >> city[ic].bi >> city[ic].ci >> city[ic].Pi >> city[ic].R;
}
visit[city[ic].ai] = 1;
search_dfs(1, 0);
if (min_cost >= MAXCOUNT)
cout << "impossible" << endl;
else
cout << min_cost << endl;
}
return 0;
}