洛谷P5340 大中锋的游乐场 题解 分层图最短路

题目链接：https://www.luogu.com.cn/problem/P5340

解题思路：

拆点/分层图最短路。

每个点 \(u\) 及到达点 \(u\) 是所吃汉堡与所喝可乐数量之差 \(p\) 对应一个二元组 \((u,p)\)。对二元组求最短路。

示例程序：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 10100;
struct Edge {
    int v, w;
    Edge() {}
    Edge(int _v, int _w) { v = _v; w = _w; }
};
vector<Edge> g[maxn];
int dis[maxn][21], T, n, m, k, a[maxn], s, t;
bool vis[maxn][21];
struct Node {
    int u, p, w;
    Node() {}
    Node(int _u, int _p, int _w) { u = _u; p = _p; w = _w; }
    bool operator < (const Node & b) const {
        return w > b.w;
    }
};
priority_queue<Node> que;

int main() {
    ios::sync_with_stdio(0);
    cin >> T;
    while (T --) {
        cin >> n >> m >> k;
        for (int i = 1; i <= n; i ++) {
            for (int j = 0; j <= 2*k; j ++) {
                dis[i][j] = -1;
                vis[i][j] = false;
            }
        }
        for (int i = 1; i <= n; i ++) {
            cin >> a[i];
            g[i].clear();
        }
        while (m --) {
            int u, v, w;
            cin >> u >> v >> w;
            g[u].push_back(Edge(v, w));
            g[v].push_back(Edge(u, w));
        }
        cin >> s >> t;
        dis[s][k+(a[s]==1 ? 1 : -1)] = 0;
        que.push(Node(s, (a[s]==1 ? 1 : -1), 0));
        while (!que.empty()) {
            Node nd = que.top();
            que.pop();
            int u = nd.u, p = nd.p, w = nd.w;
            if (vis[u][k+p]) continue;
            vis[u][k+p] = true;
            for (auto e : g[u]) {
                int v = e.v;
                int pp = p + (a[v]==1 ? 1 : -1);
                if (pp >= -k && pp <= k && (dis[v][k+pp] == -1 || dis[v][k+pp] > w + e.w)) {
                    dis[v][k+pp] = w + e.w;
                    que.push(Node(v, pp, w+e.w));
                }
            }
        }
        int ans = -1;
        for (int i = -k; i <= k; i ++) {
            if (dis[t][k+i] != -1 && (ans == -1 || ans > dis[t][k+i]))
                ans = dis[t][k+i];
        }
        cout << ans << endl;
    }
    return 0;
}