| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5481 | Accepted: 1947 |
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:
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
有n个城市m条路线，每条路线有5个值，a,b,c,p,r表示从a城到b城，如果到过c城，收费p元，否则收费r元。现在从1到n，求最小花费。
这题主要是去过的点又回来的问题，例如样例1->2->1->3->4，从2又返回了1，由于m=10，所以一个点最多访问3次(其实取2也能A，感觉还是3靠谱点）
ps：刚开始记录了map[i][j]表示i到j可以走第map[i][j]条路，因为从第i到j可以有多条路，一直wa，再改就麻烦了，换成直接枚举m条路，反正m也不大。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int inf=99999999;
//int map[15][15];
int vis[15];
int n,m;
int ans;
struct node
{
int a;
int b;
int c;
int p;
int r;
}road[15];
void dfs(int u,int v)
{
if(v>ans)
return;
if(u==n)
{
if(v<ans)
{
ans=v;
// printf(" %d\n",ans);
}
return;
}
for(int i=1;i<=m;i++)
{
if(road[i].a==u&&vis[road[i].b]<=3)
{
int b=road[i].b;
vis[b]++;
if(vis[road[i].c])
{
dfs(b,v+road[i].p);
}
else
{
dfs(b,v+road[i].r);
}
vis[b]--;
}
}
return;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(vis,0,sizeof(vis));
ans=inf;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d%d%d",&road[i].a,&road[i].b,&road[i].c,&road[i].p,&road[i].r);
}
vis[1]=1;
dfs(1,0);
if(ans==inf)
printf("impossible\n");
else
printf("%d\n",ans);
}
return 0;
}
原文地址：http://blog.csdn.net/caduca/article/details/40475953
