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：
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #include <algorithm> 5 #include <stdlib.h> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #define loop(s,i,n) for(i = s;i < n;i++) 10 #define cl(a,b) memset(a,b,sizeof(a)) 11 const int maxn = 521; 12 const int inf = 99999999; 13 using namespace std; 14 struct node 15 { 16 int u,v,len,cost; 17 int next; 18 }edges[10001]; 19 int g[105]; 20 int vis[105]; 21 int ans; 22 int n,m; 23 void dfs(int u,int val,int dis) 24 { 25 vis[u] = 1; 26 if(val < 0) 27 return; 28 if(u == n) 29 { 30 if(ans > dis) 31 ans = dis; 32 return ; 33 } 34 35 if(ans < dis) 36 return; 37 38 int i; 39 40 for(i = g[u];i != -1;i = edges[i].next) 41 { 42 int v; 43 v = edges[i].v; 44 int len,cost; 45 len = edges[i].len; 46 cost = edges[i].cost; 47 if(!vis[v]) 48 { 49 dfs(v,val-cost,dis+len); 50 vis[v]--; 51 } 52 } 53 54 } 55 56 int main() 57 { 58 int u,v,l,t; 59 int k; 60 while(~scanf("%d %d %d",&k,&n,&m)) 61 { 62 int i; 63 memset(vis,0,sizeof(vis)); 64 cl(g,-1); 65 int cnt = 0; 66 while(m--) 67 { 68 scanf("%d%d%d%d",&u,&v,&l,&t); 69 edges[cnt].u = u; 70 edges[cnt].v = v; 71 edges[cnt].len = l; 72 edges[cnt].cost = t; 73 edges[cnt].next = g[u]; 74 g[u] = cnt; 75 cnt++; 76 } 77 ans = inf; 78 dfs(1,k,0); 79 if(ans != inf) 80 printf("%d ",ans); 81 else 82 puts("-1"); 83 } 84 return 0; 85 }