【poj】【费用限制最短路】ROADs

题意

每条路有一个长度和一个花费，在花费限制内求从1
到n的最短路。

分析

只要能走到（有道路相连并且花费小于限制）就加入队列，队列中以距离为第一关键字，花费为第二关键字排序。

代码

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
#define maxn 10100
#define INF 1000000000-1
using namespace std;

struct edge{
    int to,val,cost;
    edge(int to,int val,int cost){
        this->to=to; this->val=val; this->cost=cost;
    }
};

struct state{
    int num,val,cost;//编号为num的点最小距离为val花费为cost 
};

bool operator < (const state &a,const state &b){
    if(a.val==b.val) return a.cost>b.cost;
    else return a.val>b.val;
}

typedef struct state state;
typedef struct edge edge;

int w,n,k,s,D,l,t;
int d[maxn];
vector<edge> g[maxn];


void Dij(){
    priority_queue<state> q;
    memset(d,0x3f,sizeof(d));
    int ans=INF;
    state ini;
    ini.num=1; ini.val=0; ini.cost=0;
    q.push(ini);
    while(!q.empty()){
        state cur=q.top(); q.pop();
        int curn=cur.num,curv=cur.val,curc=cur.cost;
        if(curn==n){
            ans=curv;
            break;
        }
        for(int i=0;i<g[curn].size();i++){
            int temp=g[curn][i].to;
            if(curc+g[curn][i].cost<=w){
                state p;
                p.num=temp; p.val=curv+g[curn][i].val; p.cost=curc+g[curn][i].cost;
                q.push(p);
            }
        }
    }
    if(ans<INF) printf("%d",ans);
    else printf("-1");
    return;
}

int main(){
    freopen("roads.in","r",stdin);
    freopen("roads.out","w",stdout);
    scanf("%d",&w); scanf("%d",&n); scanf("%d",&k);
    for(int i=1;i<=k;i++){
        scanf("%d%d%d%d",&s,&D,&l,&t);
        g[s].push_back(edge(D,l,t));
    }
    Dij();
    return 0;
}