poj 3411Paid Roads(走重复点的dfs )

原创于 2017-02-08 10:35:21 发布 · 304 阅读

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本文介绍了一种解决带有两种支付选项的付费道路网络中寻找从起点到终点最小成本路径的问题。通过使用深度优先搜索（DFS）算法并限制每个城市的访问次数来避免无限循环，最终实现了有效的解决方案。

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Paid Roads

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7017		Accepted: 2599

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

Source

Northeastern Europe 2002, Western Subregion

这个题目的意思是有两种付费方式。

1：如果以前经过c城市，那么就在c城市付费。。

2：如果没有经过，那么就在b城市付费。。

还有注意题目中的数据为m<10，所以每个点最多走3次。。

因为成环的话那么最少3个点，则最少3路。所以最多走3次。。。

故应用int vis[ manx]。。。

然后dfs回溯。。

#include <stdio.h>
#include <string.h>
#define INF 0x3f3f3f3f
struct node
{
    int a,b,c,p,r;
}s[20];
int vis[20];
int n, m;
int ans;
void dfs( int x, int cost)
{

    if( x == n && ans > cost)
    {
        ans = cost;return;
    }
    for( int i = 0; i < m; i++)
    {
        if(s[i].a == x &&vis[s[i].b] < 3)
        {
            vis[s[i].b]++;
            if(vis[s[i].c])
                dfs(s[i].b,cost+s[i].p);
            else dfs(s[i].b, cost+s[i].r);
            vis[s[i].b]--;
        }
    }
}
int main()
{
    scanf("%d%d", &n,&m);
    for( int i = 0; i < m; i++)
    {
        scanf("%d%d%d%d%d",&s[i].a,&s[i].b,&s[i].c,&s[i].p,&s[i].r);
    }
    memset(vis, 0,sizeof(vis));
    ans = INF;
    vis[1] = 1;
    dfs(1, 0);
    if(ans == INF)  printf("impossible\n");
    else printf("%d\n",ans);
}