[AcWing 178] 第K短路

A* 算法

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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;
typedef pair<int,PII> PIPII;

const int N = 1e6 + 10;

int n, m, S, T, K;
int h[N], rh[N], e[N], ne[N], w[N], idx;
int d[N], cnt[N];
bool st[N];

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

void dijkstra()
{
    priority_queue<PII, vector<PII>, greater<PII>> heap;
    heap.push({0, T});
    memset(d, 0x3f, sizeof d);
    d[T] = 0;
    while (heap.size()) {
        auto t = heap.top();
        heap.pop();
        int v = t.second;
        if (st[v])
            continue;
        st[v] = true;
        for (int i = rh[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});
            }
        }
    }
}

int astar()
{
    priority_queue<PIPII, vector<PIPII>, greater<PIPII>> heap;
    heap.push({d[S], {0, S}});
    while (heap.size()) {
        auto t = heap.top();
        heap.pop();
        int v = t.second.second, dist = t.second.first;
        cnt[v] ++;
        if (cnt[T] == K)
            return dist;
        for (int i = h[v]; i != -1; i = ne[i]) {
            int j = e[i];
            if (cnt[j] < K)
                heap.push({dist + w[i] + d[j], {dist + w[i], j}});
        }
    }
    return -1;
}

void solve()
{
    cin >> n >> m;
    memset(h, -1, sizeof h);
    memset(rh, -1, sizeof rh);
    for (int i = 0; i < m; i ++) {
        int a, b, c;
        cin >> a >> b >> c;
        add(h, a, b, c);
        add(rh, b, a, c);
    }
    cin >> S >> T >> K;
    if (S == T)
        K ++;
    dijkstra();
    cout << astar() << endl;
}

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

    solve();

    return 0;
}

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1. 估价函数
   每个点到终点的最短距离，$A^*$ 算法只要保证估价值小于真实值即可
2. $A^{*}$算法过程
   ① 建立反图，然后在反图上求出从终点到其他所有点的最短距离，作为每个点的估价函数
   ② 从起点开始扩展，每次取出当前的估计值最小的点，将其所有能扩展到的点全部扩展
   $估计值 = 距离起点的真实距离 + 估价值$
   ③ 当第 $k$ 次遇到终点时，就求出了从起点到终点的第 $k$ 短路