ACM: 条件最短路 poj 1724 (没剪枝…

                                                                  ROADS

Description

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

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

Input

The first line of the input contains the integer K, 0<= K <= 10000, maximum number ofcoins 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 specifyingintegers 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 anddestination cities.

Output

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

 

题意: 现在要从1 到达 n城市, 每个城市之间的链接都知道, 道路的长度和费用.

        求在小于等于费用K下最短路径.

解题思路:

               1. 条件最短路. 依然是记忆化深搜 + 剪枝.

               2. 可恨的测试数据有N多个环. TLE多次.

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 1005
const int INF = 1 << 30;

struct node
{
    int v;
    intlen;
    intcost;
    intnext;
}edges[MAX*10];

int K;
int n, m;
int next[MAX*10];
int vis[MAX];
int num;
int result;

void init()
{
    num =0;
    result =INF;
   memset(next,-1,sizeof(next));
   memset(vis,0,sizeof(vis));
}

void dfs(int v,int w,int cost)
{
    if(w> INF)
      return ;
    if(v ==n)
    {
      if(result > w)
         result = w;
      return ;
    }
    
    for(int e =next[v]; e != -1; e = edges[e].next)
    {
      int len = edges[e].len;
      int t_cost = edges[e].cost;
      if( vis[ edges[e].v ] <= 1)
      {
         vis[ edges[e].v ]++;
         if(t_cost + cost <= K&& w+len <result)
            dfs(edges[e].v,w+len,t_cost+cost);
         vis[ edges[e].v ]--;
      }
    }
}

int main()
{
//   freopen("input.txt","r",stdin);
    int u, v,len, cost;
   while(scanf("%d %d%d",&K,&n,&m) !=EOF)
    {
      num = 0;
      result = INF;
      memset(next,-1,sizeof(next));
      memset(vis,0,sizeof(vis));
      for(int i = 1; i <= m; ++i)
      {
         scanf("%d %d %d%d",&u,&v,&len,&cost);
         edges[num].v = v;
         edges[num].len = len;
         edges[num].cost = cost;
         edges[num].next = next[u];
         next[u] = num;
         num++;
      }
      vis[1] = 1;
      dfs(1,0,0);
      if(result != INF)
         printf("%d\n",result);
      else
         printf("-1\n");
    }
    return0;
}