HDU 3790 最短路径问题（双key dijkstra + 堆优化）

本文深入探讨了Dijkstra算法的增强版，通过在经典Dijkstra基础上加入费用更新机制，实现路径长度相同情况下的最小费用选择。代码示例清晰展示了算法实现细节，包括节点结构、边结构定义、初始化过程、松弛操作及优先级队列使用。

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dijs简单题，两个关键字，第二次做了，细节没处理好wa了几次

但是原理还是很简单的，就是在原来dijs算法的基础上加了个判断，松弛的时候当路径长度相同时更新最小费用

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#define ll long long
#define mod 1000000007
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 1005;
const int maxm = 100005;
int n, m;
int d[maxn], d1[maxn];
bool vis[maxn];
int head[maxn];
int cnt;
struct edge
{
    int to, next, w, w1;
}e[maxm];
struct node
{
    int v, c, c1;
    friend bool operator < (const node &a, const node &b)
    {
        if(a.c == b.c)
            return a.c1 > b.c1;
        else 
            return a.c > b.c;
    }
    node(int x = 0, int y = 0, int z = 0) : v(x), c(y), c1(z) {}
};
void addedge(int u, int v, int w, int w1)
{
    e[cnt].w = w;
    e[cnt].w1 = w1;
    e[cnt].to = v;
    e[cnt].next = head[u];
    head[u] = cnt ++;
}
void init()
{
    cnt = 0;
    memset(head, -1, sizeof(head));
    memset(d, 0x3f, sizeof(d));
    memset(d1, 0x3f, sizeof(d1));
    memset(vis, 0, sizeof(vis));
}
void dijs(int s)
{
    priority_queue<node> q;
    d1[s] = d[s] = 0;
    node h;
    q.push(node(s, 0, 0));
    while(! q.empty())
    {
        h = q.top();
        q.pop();
        int u = h.v;
        if(vis[u]) continue;
        vis[u] = 1;
        for(int i = head[u]; ~ i; i = e[i].next)
        {
            int v = e[i].to;
            int w = e[i].w;
            int w1 = e[i].w1;
            if(vis[v]) continue;
            if(d[v] > d[u] + w)
            {
                d[v] = d[u] + w;
                d1[v] = d1[u] + w1;
                q.push(node(v, d[v], d1[v]));
            }
            else if(d[v] == d[u] + w && d1[v] > d1[u] + w1)
            {
                d[v] = d[u] + w;
                d1[v] = d1[u] + w1;
                q.push(node(v, d[v], d1[v]));
            }
        }
    }
}
int main()
{
    while(scanf("%d%d", &n, &m) != EOF && (n + m))
    {
        init();
        int s, t;
        int u, v, len, cost;
        for(int i = 1; i <= m; i ++)
        {
            scanf("%d%d%d%d", &u, &v, &len, &cost);
            addedge(u, v, len, cost);
            addedge(v, u, len, cost);
        }
        scanf("%d%d", &s, &t);
        dijs(s);
        printf("%d %d\n", d[t], d1[t]);
    }
    return 0;
}