poj 1724 ROADS 很水的dfs

题意：给你N个城市和M条路和K块钱，每条路有话费，问你从1走到N的在K块钱内所能走的最短距离是多少

链接：http://poj.org/problem?id=1724

直接dfs搜一遍就是

代码：

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <stdlib.h>
 #include <vector>
 #include <queue>
 #include <stack>
 #define loop(s,i,n) for(i = s;i < n;i++)
 #define cl(a,b) memset(a,b,sizeof(a))
 const int maxn = ;
 const int inf = ;
 using namespace std;
 struct node
 {
     int u,v,len,cost;
     int next;
 }edges[];
 int g[];
 int vis[];
 int ans;
 int n,m;
 void dfs(int u,int val,int dis)
 {
     vis[u] = ;
     if(val < )
     return;
     if(u == n)
     {
         if(ans > dis)
         ans = dis;
         return ;
     }

     if(ans < dis)
     return;

     int i;

     for(i = g[u];i != -;i = edges[i].next)
     {
         int v;
         v = edges[i].v;
         int len,cost;
         len = edges[i].len;
         cost = edges[i].cost;
         if(!vis[v])
         {
             dfs(v,val-cost,dis+len);
             vis[v]--;
         }
     }

 }

 int main()
 {
     int u,v,l,t;
     int k;
     while(~scanf("%d %d %d",&k,&n,&m))
     {
         int i;
         memset(vis,,sizeof(vis));
         cl(g,-);
         int cnt = ;
         while(m--)
         {
             scanf("%d%d%d%d",&u,&v,&l,&t);
             edges[cnt].u = u;
             edges[cnt].v = v;
             edges[cnt].len = l;
             edges[cnt].cost = t;
             edges[cnt].next = g[u];
             g[u] = cnt;
             cnt++;
         }
         ans = inf;
         dfs(,k,);
         if(ans != inf)
         printf("%d\n",ans);
         else
         puts("-1");
     }
     return ;
 }