【牛客网】KY152 最短路径问题

这道题是求最短路径，由于n最大可以为1000，而Floyd算法的复杂度为O(n³)，因此会超时，所以只能用Dijkstra算法，题目要求我们不仅求出最短路径，还要求出最短路径下的最少花费：

#include <iostream>
#include <vector>
#include <queue>
#include <climits>
#include <cstring>

using namespace std;

const int N = 1000 + 10;
const int INF = INT_MAX;

struct Edge{
    int to, length, price;
    Edge(int t, int l, int p): to(t), length(l), price(p){}
};

struct Point{
    int number, distance;
    Point(int n, int d): number(n), distance(d){}
    bool operator<(const Point p) const{
        return distance > p.distance;
    }
};

vector<Edge> graph[N];
int dis[N], cost[N];

void Dijkstra(int s){
    priority_queue<Point> pq;
    dis[s] = 0;
    cost[s] = 0;
    pq.push(Point(s, dis[s]));
    while(!pq.empty()){
        int u = pq.top().number;
        pq.pop();
        for(int i = 0; i < graph[u].size(); i++){
            int v = graph[u][i].to;
            int d = graph[u][i].length;
            int p = graph[u][i].price;
            if((dis[v] == dis[u] + d && cost[v] > cost[u] + p) || (dis[v] > dis[u] + d)){
                dis[v] = dis[u] + d;
                cost[v] = cost[u] + p;
                pq.push(Point(v, dis[v]));
            }
        }
    }
}

int main(){
    int n, m, a, b, d, p, s, t;
    while(scanf("%d%d", &n, &m) != EOF && n && m){
        memset(graph, 0, sizeof(graph));
        fill(dis, dis + n + 1, INF);
        fill(dis, dis + n + 1, INF);
        for(int i = 0; i < m; i++){
            scanf("%d%d%d%d", &a, &b, &d, &p);
            graph[a].push_back(Edge(b, d, p));
            graph[b].push_back(Edge(a, d, p));
        }
        scanf("%d%d", &s, &t);
        Dijkstra(s);
        printf("%d %d\n", dis[t], cost[t]);
    }
    return 0;
}