图论三种做法：朴素版Dijkstra、堆优化（优先队列）Dijkstra、spfa（队列优化版Bellman-Ford）

【模板】单源最短路径

题目传送门：P3371 【模板】单源最短路径（弱）

//1、朴素版Dijkstra
#include <iostream>
#include <cstring>
using namespace std;
const int N = 2001, INF = 0x3f3f3f3f;

int n, m, S, dist[N], d[N][N];
bool st[N];

void dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    dist[S] = 0;
    for(int i = 0; i < n; i ++)
    {
        int t = -1;
        for(int j = 1; j <= n; j ++)
            if(!st[j] && (t == -1 || dist[t] > dist[j]))
                t = j;
        st[t] = true;

        for(int j = 1; j <= n; j ++)
            if(dist[j] > dist[t] + d[t][j])
                dist[j] = dist[t] + d[t][j];
    }
}

int main()
{
    cin >> n >> m >> S;
    memset(d, 0x3f, sizeof d);//注意两点之间的值也要初始化，不然得到的结果全部为0

    while(m --)
    {
        int a, b, c;
        cin >> a >> b >> c;
        d[a][b] = min(d[a][b], c);
    }

    dijkstra();

    for(int i = 1; i <= n; i ++)
    {
        if(dist[i] == 0x3f3f3f3f) cout << 2147483647 << ' '; // 2147483647 = (1 << 31) - 1;
        else cout << dist[i] << ' ';
    }
    cout << endl;

    return 0;
}

//2、堆优化版Dijkstra(优先队列)
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int N = 10001, M = 5e5 + 10, INF = 0x3f3f3f3f;
typedef pair<int, int> PII;
int n, m, S, d[N], e[M], ne[M], h[N], w[M], idx;
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

void dijkstra()
{
    memset(d, 0x3f, sizeof d);
    d[S] = 0;
    priority_queue<PII, vector<PII>, greater<PII> > q;
    q.push({0, S});
    while(q.size())
    {
        PII t = q.top();
        q.pop();
        int ver = t.second, distance = t.first;
        if(st[ver]) continue;
        st[ver] = true;

        for(int i = h[ver]; ~i; i = ne[i])
        {
            int j = e[i];
            if(d[j] > distance + w[i]) 
            {
                d[j] = distance + w[i];
                if(!st[j])
                {
                    q.push({d[j], j});
                }
            }
        }
    }
}

int main()
{
    cin >> n >> m >> S;
    memset(h, -1, sizeof h);

    while(m --)
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }

    dijkstra();

    for(int i = 1; i <= n; i ++)
    {
        if(d[i] == 0x3f3f3f3f) cout << 2147483647 << ' ';
        else cout << d[i] << ' ';
    }
    cout << endl;

    return 0;
}

//3、spfa：队列优化版bellman-ford
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int N = 10001, M = 5e5 + 10, INF = 0x3f3f3f3f;
int n, m, S, d[N], e[M], ne[M], h[N], w[M], idx;
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

void spfa()
{
    memset(d, INF, sizeof d); d[S] = 0;
    queue<int> q; q.push(S); st[S] = true;
    while(q.size())
    {
        int t = q.front();
        q.pop();
        st[t] = false; //spfa可以入队多次，所以入队设为true，出队设为false

        for(int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if(d[j] > d[t] + w[i]) //注意不要用d[j] = min(d[j], d[t] + w[i]),会超时，if判断比min、max操作快。
            {
                d[j] = d[t] + w[i];
                if(!st[j])
                {
                    q.push(j);
                    st[j] = true;
                }
                
            }
        }
    }
}

int main()
{
    cin >> n >> m >> S;
    memset(h, -1,sizeof h);

    while(m --)
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }

    spfa();

    for(int i = 1; i <= n; i ++)
    {
        if(d[i] == 0x3f3f3f3f) cout << 2147483647 << ' ';
        else cout << d[i] << ' ';
    }
    cout << endl;

    return 0;
}

 注意有些题目可能会卡spfa，因为时间复杂度是O(k * m)，最坏情况下会变为O(n * m)。

所以最保险的做法还是堆优化（优先队列）的dijkstra

以下代码可以同时AC  P3371 【模板】单源最短路径（弱化版）和  P4779 【模板】单源最短路径（标准版）

//堆优化dijkstra
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
typedef pair<int, int> PII;
const int N = 1e5 + 10, M = 5e5 + 10, INF = 0x3f3f3f3f;
int n, m, S, e[M], ne[M], h[N], w[M], idx, d[N];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx ++;
}

void dijkstra()
{
    memset(d, 0x3f, sizeof d);
    priority_queue<PII, vector<PII>, greater<PII> > q;
    d[S] = 0;
    q.push({0, S});
    while(q.size())
    {
        PII t = q.top();
        q.pop();

        int ver = t.second, distance = t.first;
        if(st[ver]) continue; //dijstra每个点只走一次，所出队后设为true，不再走了
        st[ver] = true;
        for(int i = h[ver]; ~i; i = ne[i])
        {
            int j = e[i];
            if(d[j] > distance + w[i])
            {
                d[j] = distance + w[i];
                if(!st[j]) //走到的j点是不是没有走过
                {
                    q.push({d[j], j});
                }
            }
        }
    }
}

int main()
{
    cin >> n >> m >> S;
    memset(h, -1, sizeof h);
    while(m --)
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }

    dijkstra();

    for(int i = 1; i <= n; i ++)
    {
        if(d[i] == 0x3f3f3f3f) cout << 2147483647 << ' ';
        else cout << d[i] << ' ';
    }
}