poj 1724(最短路+优先队列)

ROADS

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13436		Accepted: 4921

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

题意:在规定的花费内从 1 - > n 的最短路。
题解:这题有很多方法,但是优先队列+dijstra 是最快的的。spfa+dp，dfs都可行。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
using namespace std;
const int N = ;
int K,n,m;
struct Node{
    int u,len,cost;
};
bool operator < (Node a,Node b){
    if(a.len==b.len) return a.cost > b.cost;
    return a.len > b.len;
}
struct Edge{
    int v,w,cost,next;
}edge[N*N];
int head[N];
int tot;
void init(){
    memset(head,-,sizeof(head));
    tot = ;
}
void addEdge(int u,int v,int w,int cost,int &k){
    edge[k].v = v,edge[k].cost = cost,edge[k].w = w,edge[k].next = head[u],head[u] = k++;
}
int bfs(){
    priority_queue <Node > q;
    Node s;
    s.u = ,s.len = ,s.cost = ;
    q.push(s);
    while(!q.empty()){
        Node now = q.top();
        q.pop();
        if(now.u == n) return now.len;
        for(int k=head[now.u];k!=-;k=edge[k].next){
            int v = edge[k].v,cost = edge[k].cost,len = edge[k].w;
            if(now.cost+cost>K) continue;
            Node next;
            next.u = v;
            next.cost = now.cost+cost;
            next.len = now.len+len;
            q.push(next);
        }
    }
    return -;
}
int main()
{
        init();
        scanf("%d",&K);
        scanf("%d%d",&n,&m);
        for(int i=;i<=m;i++){
            int u,v,w,cost;
            scanf("%d%d%d%d",&u,&v,&w,&cost);
            addEdge(u,v,w,cost,tot);
        }
        printf("%d\n",bfs());
        return ;
}