Dijkstra嘛,就是每次从最短路未固定的点中找到已知最短路最短的点,然后将它固定,并更新这个点连接的其他点的最短路。最开始时,源点到源点的最短路为0。
所以,复习了一遍Dijkstra然后发现了几个函数

  • make_heap (first, last, comp) : 把一个数组搞成一个堆
  • push_heap (first, last, comp) : 让数组末尾的数浮到堆中正确的位置
  • pop_heap (first, last, comp) :把堆gui头丢到数组末尾

那么简单地封装一下就是

void push (int x) {
    heap[++hsize] = x;
    std::push_heap(heap + 1, heap + 1 + hsize, cmp1);
}
void pop (void) {
    std::pop_heap(heap + 1,heap + 1 + hsize--, cmp1);
}

用数组实现的堆跑得比香港记者还快,vector实现的优先队列不知道高到哪里去了。
如图

模板题代码如下

#include <cstdio>
#include <algorithm>

const int MAXM = 500000 + 5, MAXN = 100000 + 5, INF = 2147483647;

int n, m, s;

struct ed {
    int to, nex, w;
} e[MAXM];

int head[MAXN], dis[MAXN], hsize;
bool v[MAXN];
int newp;

struct node {
    int id, v;
} heap[MAXN];

void insert (int p1, int p2, int w) {
    ++newp;
    e[newp].to = p2;
    e[newp].w = w;
    e[newp].nex = head[p1];
    head[p1] = newp;
}

bool cmp1 (node x, node y) {
    return x.v > y.v;
}

void push (node x) {
    heap[++hsize] = x;
    std::push_heap(heap + 1, heap + 1 + hsize, cmp1);
}

void pop (void) {
    std::pop_heap(heap + 1,heap + 1 + hsize--, cmp1);
}

void dij (int s) {
    for (int i = 1; i <= n; ++i) {
        dis[i] = INF;
        v[i] = 0;
    }
    dis[s] = 0;
    hsize = 0;
    push((node){s, 0});
    
    while (hsize) {
        node u = heap[1];
        pop();
        if (v[u.id]) continue;//已固定的点 
        v[u.id] = 1;
        for (int i = head[u.id]; i; i = e[i].nex) {
            int y = e[i].to;
            if (dis[y] > u.v + e[i].w) {
                dis[y] = u.v + e[i].w;
                push((node){y, dis[y]});
            }
        }
    }
}

int main (void) {
    scanf("%d%d%d", &n, &m, &s);
    
    for (int i = 1; i <= n; ++i) {
        head[i] = 0;
        heap[i] = (node){0, 0};
    }
    
    for (int i = 1; i <= m; ++i) {
        e[i] = (ed){0, 0, 0};
    }
    
    {
        int p1, p2, w;
        for (int i = 1; i <= m; ++i) {
            scanf("%d%d%d", &p1, &p2, &w);
            insert(p1, p2, w);
        }
    }
    
    dij(s);
    
    for (int i = 1; i <= n; ++i) {
        printf("%d ", dis[i]);
    }
    putchar('\n');
    
    return 0;
}