[ZJOI2006]物流运输

题目传送门

这道题的思路应该为$dp$+最短路。

状态设计：

$g[i][j]$表示从第$i$时刻（注意，是时刻）到第$j$时刻过程中不改变路线时的最优解。显然，是将在这期间内所有要关闭的港口从图中删除，然后求最短路。最后乘上$j-i$即可。

$f[i]$表示第$i$时刻的最优解。

状态转移：设之前的第$j$时刻，从第$j$时刻转移到第$i$时刻，要取$f[j]+g[j][i]+K$的最小值。

所以这题得解，时间复杂度为$O(n^2mlogm)$，鉴于本题$n leq 100$、$m leq 20$，该复杂度无压力。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b) for (register int i = a; i <= b; ++i)
#define minn(a, b) a = min(a, b);
#define LL long long

inline int read() {
    int w = 0, f = 1; char c = getchar();
    while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();
    while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();
    return w * f;
}

const int maxn = 20 + 5, maxm = 100 + 5;

struct Edge {
    int u, v, w, pre;
};

struct Node {
    int u, d;
    bool operator < (const Node &rhs) const {
        return d > rhs.d;
    }
};

int n, m, K, e, d, ban[maxn], b[maxm][maxm];
LL f[maxm], g[maxm][maxm];

priority_queue<Node> q, empty;

struct Graph {
    Edge edges[maxn * maxn << 1];
    int n, m;
    int G[maxn];
    int dis[maxn], vis[maxn];
    void init(int n) {
        this->n = n;
        m = 0;
        memset(G, 0, sizeof(G));
    }
    void Add(int u, int v, int w) {
        edges[++m] = (Edge){u, v, w, G[u]};
        G[u] = m;
    }
    int dijkstra() {
        memset(dis, 0x3f, sizeof(dis));
        rep(i, 1, n) vis[i] = ban[i];
        q = empty;
        q.push((Node){1, 0});
        dis[1] = 0;
        while (!q.empty()) {
            int u = q.top().u; q.pop();
            if (vis[u]) continue;
            vis[u] = 1;
            for (register int i = G[u]; i; i = edges[i].pre) {
                Edge &e = edges[i];
                if (dis[u] + e.w < dis[e.v]) {
                    dis[e.v] = dis[u] + e.w;
                    q.push((Node){e.v, dis[e.v]});
                }
            }
        }
        return dis[n];
    }
} G;

int main() {
    n = read(), m = read(), K = read(), e = read();
    G.init(m);
    
    rep(i, 1, e) {
        int u = read(), v = read(), w = read();
        G.Add(u, v, w);
        G.Add(v, u, w);
    }

    memset(b, 0, sizeof(b));

    d = read();
    rep(i, 1, d) {
        int P = read(), s = read(), t = read();
        rep(x, s, t) b[P][x] = 1;
    }

    rep(i, 0, n) {
        memset(ban, 0, sizeof(ban));
        rep(j, i+1, n) {
            rep(k, 1, m) ban[k] |= b[k][j];
            g[i][j] = (LL)G.dijkstra() * (j - i);
        }
    }

    memset(f, 0x3f, sizeof(f));
    f[0] = -K;
    rep(i, 1, n)
        rep(j, 0, i-1)
            minn(f[i], f[j] + g[j][i] + K);
    f[0] = 0;

    printf("%lld", f[n]);

    return 0;
}