stonechina0616

描述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. 
输入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.输出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. 
样例输入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 样例输出11

题目大意:每一条边都有两个参数：距离和花费，寻找从1到n的最短路，满足花费小于k
解法：dijkstra+priority_queue不需要记录dis[]数组，而是将边处理后入堆，入堆的元素包含三个参数：点的编号、到这个点的总距离与总花费，当出堆元素点的编号为n时，到这个点的总距离就是总最小距离。
代码：#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<vector>#include<queue>#define rep(q,p) for(int q=1;q<=p;q++)using namespace std;const int maxn=105;int n,p,m;struct hehe{ int point,num,coin; bool operator < (const hehe & hehehe) const { if (num==hehehe.num) return hehehe.point<point; else return hehehe.num<num; }};vector<hehe>v[maxn];priority_queue<hehe>q;
int dijkstra(int st,int en){ hehe tmp; tmp.point=st,tmp.num=0,tmp.coin=0; q.push(tmp); while (!q.empty()) { tmp=q.top(); q.pop(); int k=tmp.point; if (k==en) return tmp.num; for (int i=0;i<v[k].size();i++) { int tp=v[k][i].point,tnum=v[k][i].num,tcoin=v[k][i].coin; if (tcoin+tmp.coin<=p) { hehe ts; ts.point=tp; ts.num=tmp.num+tnum; ts.coin=tcoin+tmp.coin; q.push(ts); } } } return -1;}
int main(){ freopen("in.txt","r",stdin); scanf("%d%d%d",&p,&n,&m); rep(i,m) { int st,en,di,ci; scanf("%d%d%d%d",&st,&en,&di,&ci); hehe tmp; tmp.point=en; tmp.num=di; tmp.coin=ci; v[st].push_back(tmp); } printf("%d\n",dijkstra(1,n)); return 0;}