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起点到终点的最短距离

堆优化 dijkstra

复杂度 \(O(m \cdot log(n)) = 6200 \times log(2500) \approx 7 \times 10^4\)

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点击查看代码

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int,int> PII;

const int N = 1e6 + 10;

int n, m, sp, ep;
int h[N], e[N], ne[N], w[N], idx;
int d[N];
bool st[N];

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

void dijkstra(int sp)
{
    memset(d, 0x3f, sizeof d);
    memset(st, false, sizeof st);
    priority_queue<PII, vector<PII>, greater<PII>> heap;
    heap.push({0, sp});
    d[sp] = 0;
    while (heap.size()) {
        auto t = heap.top();
        heap.pop();
        auto v = t.second;
        if (st[v])
            continue;
        st[v] = true;
        for (int i = h[v]; i != -1; i = ne[i]) {
            int j = e[i];
            if (d[j] > d[v] + w[i]) {
                d[j] = d[v] + w[i];
                heap.push({d[j], j});
            }
        }
    }
}

void solve()
{
    cin >> n >> m >> sp >> ep;
    memset(h, -1, sizeof h);
    for (int i = 0; i < m; i ++) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
        add(b, a, c);
    }
    dijkstra(sp);
    cout << d[ep] << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    solve();

    return 0;
}

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1. 非负权图尽量不用 \(SPFA\)，容易被卡，用堆优化 \(dijkstra\)
2. 此题的堆优化 \(dijkstra\) 模板可求以 \(sp\) 作为起点，到其他所有点的最短路