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
n城市数,m道路数
u~v,c如果经过c则用为p否则为r
爆搜,注意恶性递增的环,控制点访问次数不大于3
#include <iostream>
#include <string.h>
using namespace std;
int n,m; //城市数,道路数
int vis[11]; //记录城市的访问次数,当点访问数大于等于4时退出该层dfs,然后继续搜索(防止恶性增加的环)
int mmin; //最小总花费
struct haha{
int u,v;
int c;
int p,r;
}we[11]; //u~v,经过c时费用为p否则为r
void dfs(int s,int minn){
if(s==n && mmin>minn){
mmin=minn;
return;
}
for(int i=1;i<=m;i++){
if(s==we[i].u && vis[we[i].v]<=3){//控制点访问不超过3次
int v=we[i].v;//下个点
vis[v]++;
if(vis[ we[i].c ])
dfs(v,minn+we[i].p);
else
dfs(v,minn+we[i].r);
vis[v]--;
}
}
return;
}
int main(void){
while(cin>>n>>m){
memset(vis,0,sizeof(vis));
mmin=99999;
for(int i=1;i<=m;i++){
scanf("%d %d %d %d %d",&we[i].u,&we[i].v,&we[i].c,&we[i].p,&we[i].r);
}
vis[1]=1;
dfs(1,0);
if(mmin==99999){
printf("impossible\n");
}
else{
printf("%d\n",mmin);
}
}
return 0;
}