看不出来有什么差别，但是一个 16pts，一个 100pts。题目是单源最短路径。

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

using namespace std;

const int N = 100010;
const int M = 200010;

struct edge {
    int v, w;
};

int n, m, s;
int dis[N];
vector<edge> map[N];
bool vis[N];

void dijsktra() {
    // queue<pair<int, int> > q;
    priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > q;
    memset(dis, 127, sizeof dis);
    q.push(make_pair(dis[s] = 0, s));
    while (q.size()) {
        auto it = q.top(); q.pop();
        int w = it.first, u = it.second;
        if (vis[u]) continue;
        vis[u] = true;
        for (auto it : map[u]) 
            if (dis[it.v] > dis[u] + it.w) q.push(make_pair(dis[it.v] = dis[u] + it.w, it.v));
    }
}

int main() {
    cin >> n >> m >> s;
    for (int i = 1; i <= m; i++) {
        int u, v, w; cin >> u >> v >> w;
        map[u].push_back((edge){ v, w });
    }
    dijsktra();
    for (int i = 1; i <= n; i++) cout << dis[i] << " ";
    puts("");
    return 0;
}

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

using namespace std;

const int N = 100010;
const int M = 200010;

struct edge {
    int v, w;
};

int n, m, s;
int dis[N];
vector<edge> map[N];
bool vis[N];

void dijsktra() {
    queue<pair<int, int> > q;
    memset(dis, 127, sizeof dis);
    q.push(make_pair(dis[s] = 0, s));
    while (q.size()) {
        auto it = q.front(); q.pop();
        int w = -it.first, u = it.second;
        if (vis[u]) continue;
        vis[u] = true;
        for (auto it : map[u]) 
            if (dis[it.v] > dis[u] + it.w) q.push(make_pair(-(dis[it.v] = dis[u] + it.w), it.v));
    }
}

int main() {
    cin >> n >> m >> s;
    for (int i = 1; i <= m; i++) {
        int u, v, w; cin >> u >> v >> w;
        map[u].push_back((edge){ v, w });
    }
    dijsktra();
    for (int i = 1; i <= n; i++) cout << dis[i] << " ";
    puts("");
    return 0;
}

利用负值来达到用 priority_queue 实现从小到大排序的技巧 WA 了，但是直接用 priority_queue<greater<>> 这样的却 AC 了