poj1724

ROADS

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10804		Accepted: 3976

Description

N cities named with numbers 1 ... N are connected with one-way roads. Each road has two parameters associated with it : the road length and the toll that needs to be paid for the road (expressed in the number of coins).
Bob and Alice used to live in the city 1. After noticing that Alice was cheating in the card game they liked to play, Bob broke up with her and decided to move away - to the city N. He wants to get there as quickly as possible, but he is short on cash.

We want to help Bob to find the shortest path from the city 1 to the city N that he can afford with the amount of money he has.

Input

The first line of the input contains the integer K, 0 <= K <= 10000, maximum number of coins that Bob can spend on his way.
The second line contains the integer N, 2 <= N <= 100, the total number of cities.

The third line contains the integer R, 1 <= R <= 10000, the total number of roads.

Each of the following R lines describes one road by specifying integers S, D, L and T separated by single blank characters :

• S is the source city, 1 <= S <= N
• D is the destination city, 1 <= D <= N
• L is the road length, 1 <= L <= 100
• T is the toll (expressed in the number of coins), 0 <= T <=100

Notice that different roads may have the same source and destination cities.

Output

The first and the only line of the output should contain the total length of the shortest path from the city 1 to the city N whose total toll is less than or equal K coins.
If such path does not exist, only number -1 should be written to the output.

Sample Input

5
6
7
1 2 2 3
2 4 3 3
3 4 2 4
1 3 4 1
4 6 2 1
3 5 2 0
5 4 3 2

Sample Output

11

Source

CEOI 1998

 

 1 //邻接表存储结构+剪枝,优化时间复杂度
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 using namespace std;
 6 
 7 const int MAXWAY = 10010;
 8 const int MAXCITY = 110;
 9 const int INF = 0xffffff0;
10 struct Node{
11     int s,d,l,t;
12     int next;      //邻接表的表头指针
13 }Node[MAXWAY];
14 int k,n,r;
15 int totallen;   //存储当前路径中的路径长度
16 int totalcost; //存储当前路径中需要的话费
17 int minLen;    //存储当前情况下最短的路径长度
18 int visited[110]; //存储是否访问过某个城市
19 
20 int head[MAXWAY];   //存储链表信息
21 
22 void DFS(int i)
23 {
24     if(i==n)
25     {
26         minLen = min(minLen,totallen);
27         return ;
28     }
29     else
30     {
31         for(int j=head[i];j!=-1;j=Node[j].next)   //遍历 以i为起点的其他的所有
32         {
33             if(!visited[Node[j].d])
34             {
35                 if(totalcost+Node[j].t>k)  //如果加上当前的这条路的费用超过了k，则跳过
36                     continue;
37                 if(totallen+Node[j].l>minLen)  //如果在费用没有超过的情况下加上该条路的长度，超过了当前的最短长度，则跳过
38                     continue;
39                 totallen = totallen + Node[j].l;
40                 totalcost = totalcost + Node[j].t;
41                 visited[Node[j].d] = 1;
42                 DFS(Node[j].d);
43                 visited[Node[j].d] = 0;
44                 totallen -= Node[j].l;
45                 totalcost -= Node[j].t;
46 
47             }
48         }
49     }
50 }
51 
52 int main()
53 {
54     while(scanf("%d%d%d",&k,&n,&r)!=EOF)
55     {
56         memset(head,-1,sizeof(head));
57         memset(visited,0,sizeof(visited));
58         for(int i=0;i<r;i++)    //接受数据同时，利用头插法建立邻接表
59         {
60             scanf("%d%d%d%d",&Node[i].s,&Node[i].d,&Node[i].l,&Node[i].t);
61             Node[i].next = head[Node[i].s];      //如果当前不存在元素，则此时的Node[i]就是尾节点，它的next是-1
62             //否则，Node[i].next就是指向当前的链表最头部的那个Node元素。
63             head[Node[i].s] = i;   //修改当前的head[Node[i].s]的指向，使head[i]的值始终指向该条链表的头部元素，以便执行头插法
64         }
65         totallen = 0;
66         totalcost = 0;
67         minLen = INF;
68         DFS(1);
69         if(minLen<INF)
70             printf("%d\n",minLen);
71         else
72             printf("-1\n");
73     }
74     return 0;
75 }

 

转载于:https://www.cnblogs.com/jhldreams/p/3911896.html