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题面：

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.

题目大意：

一共有N个城市，分别命名为1-N。通过单向道路连接。每条道路有它的长度和走这条道路了需要的钱数。

Bob想从1走到N。手中有一定数量的钱。问最短的可行路径是多长。

大致思路：

用邻接表对图进行储存。

深度优先搜索。纯搜索会爆时间，所以需要进行剪枝。

1. 当当下行走的道路长度已经大于等于已经成功走过的道路长度时，不必继续搜索。
2. 当走这条路的花费加上已有的花费大于总钱数时，不必继续搜索。
3. 将每次走过某一点的花费值所走的长度进行记录。当再次走过这一点时，如果长度大于所记录的长度，不必继续搜索。

代码：

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        using namespace std;typedef struct {    int end;    int len,spend;}Road;vector< vector 
        
          > G(110);//相当于二维数组int k,n,r;int minlen,totallen,totalcost;int visit[110];int minl[110][10010];void dfs(int s){ if(s==n){ minlen= min (minlen,totallen); return ; } for(int i=0;i
         
           k)//剪枝2 continue; if(!visit[r.end]){ if(totallen +r.len >=minlen)//剪枝1 continue; if( totallen +r.len >= minl[r.end][totalcost +r.spend])//剪枝3 continue; minl[r.end][totalcost + r.spend] =totallen +r.len;//剪枝3的记录 totallen += r.len; totalcost +=r.spend; visit[r.end] = 1; dfs(r.end); visit[r.end] = 0; totallen -= r.len;//消除这一次搜索的影响 totalcost -=r.spend; } }}int main(){ cin>>k>>n>>r; for(int i=0;i
          
           >s>>r.end>>r.len>>r.spend; if(s != r.end) G[s].push_back(r); } memset(visit,0,sizeof(visit)); for(int i=0;i<110;++i) for(int j=0;j<10010;++j) minl[i][j]= 1<<30; totallen=0; minlen=1 << 30;//利用位运算变成一个极大值，上同理 totalcost=0; visit[1]=1; dfs(1); if(minlen< (1 << 30)){ cout<
           
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