poj3411Paid Roads

题目链接：

http://poj.org/problem?id=3411

题目：

Paid Roads

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5080 Accepted: 1784

Description

A network of m roads connects N cities (numbered from 1 toN). 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 roadi from city a_i to city b_i:

• in advance, in a city c_i (which may or may not be the same asa_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 ofa_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 cityN. If the trip is not possible for any reason, the line must contain the word ‘impossible’.

Sample Input

4 51 2 1 10 102 3 1 30 503 4 3 80 802 1 2 10 101 3 2 10 50

Sample Output

110

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。。。

所以代码为：

#include<iostream>#include<cstdio>#include<cstring>#define INF 0x3f3f3f3fconst int maxn=10+10;int vis[maxn];int min_cost,fee;int n,m;struct node{    int a,b,c,p,r;}e[maxn];void dfs(int x,int fee){    if(x==n&&min_cost>fee)    {        min_cost=fee;        return;    }    if(x==n)        return;    for(int i=1;i<=m;i++)    {        if(e[i].a==x&&vis[e[i].b]<3)        {            vis[e[i].b]++;            if(vis[e[i].c])                dfs(e[i].b,fee+e[i].p);            else                dfs(e[i].b,fee+e[i].r);            vis[e[i].b]--;        }    }}int main(){   while(scanf("%d %d",&n,&m)!=EOF)   {       memset(vis,0,sizeof(vis));       min_cost=INF;       for(int i=1;i<=m;i++)       {           scanf("%d%d%d%d%d",&e[i].a,&e[i].b,&e[i].c,&e[i].p,&e[i].r);       }       vis[1]=1;       dfs(1,0);       if(min_cost!=INF)          printf("%d\n",min_cost);       else         printf("impossible\n");   }   return 0;}