ROADS

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 10742	Accepted: 3949

Description

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. 

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

解決方案：此題我是這樣做的，用上優先佇列，在費用可行的情況下，不斷鬆弛路徑。其實也相當於bfs+優先佇列。首先路徑最短的優先順序最高，其次是花費，通過不斷的把符合費用要求能到達的點加入優先佇列，每次出隊即更新能到達的點。最後如果出隊的點是N，演算法結束，得到的路徑既是在花費符合的情況下最短的，這題考察的是能不能深刻理解dijkstra的原理，並運用。

code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define MMAX 10003
#define Max 103
using namespace std;
int K,N,R,k;
int head[Max];
struct edge
{

    int from,to,len,cost;
    int next;
} E[MMAX];
struct stay
{

    int dis,cost,x;
    bool operator<(const stay &s)const
    {
        if(dis!=s.dis)
        {
            return dis>s.dis;
        }
        else return cost>s.cost;

    }

};
void add(int from,int to,int len,int cost)
{
    E[k].from=from;
    E[k].to=to;
    E[k].len=len;
    E[k].cost=cost;
    E[k].next=head[from];
    head[from]=k++;

}
int dijkstra()
{
    priority_queue<stay> Q;
    stay in;
    in.dis=0,in.cost=0,in.x=1;
    Q.push(in);
    while(!Q.empty())
    {
        stay out=Q.top();

        if(out.x==N) {return out.dis;}
        Q.pop();
        for(int v=head[out.x]; v!=-1; v=E[v].next)
        {
            if(out.cost+E[v].cost<=K)
            {
               stay temp;
               temp.x=E[v].to;
               temp.dis=out.dis+E[v].len;
               temp.cost=out.cost+E[v].cost;
               Q.push(temp);
            }
        }
    }

}
int main()
{
    while(~scanf("%d%d%d",&K,&N,&R))
    {
        memset(head,-1,sizeof(head));
        k=0;
        int from,to,len,cost;
        for(int i=0; i<R; i++)
        {
            scanf("%d%d%d%d",&from,&to,&len,&cost);
            add(from,to,len,cost);
        }
        printf("%d\n",dijkstra());

    }
    return 0;
}