[BZOJ 1266][AHOI 2006] 上学路线

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题目大意

题目有两问。第一问是求给定的两点间最短路；第二问中，每条边除了权值 \(w\) 以外还有一个代价 \(c\)，求使那两点间最短路变大（不再联通也算）的最小代价。

\(2 \leqslant n \leqslant 500\)

\(1 \leqslant m \leqslant 124,750\)

\(1 \leqslant w, \; c \leqslant 10,000\)

题目链接

BZOJ 1266

题解

第一问直接跑 Dijkstra 就行了。。。

第二问，对于所有在最短路上的边，以代价为容量建最小割；判断一条边是否在最短路上，以起点、终点为源各跑一遍 Dijkstra，满足 \(e.u.dist[s] + e.w + e.v.dist[t] = t.dist[s] = s.dist[t]\) 就是。

代码

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#include <cstdio>
#include <climits>
#include <queue>
#include <algorithm>
#include <new>
const int MAXN = 505;
const int MAXM = 124755;
struct EdgeD;
struct NodeD {
    EdgeD *e;
    int dist[2], id;
    bool vis[2];
} ND[MAXN];
struct EdgeD {
    NodeD *u, *v;
    EdgeD *next;
    int w, c;
    EdgeD() {}
    EdgeD(NodeD *u, NodeD *v, int w, int c) : u(u), v(v), w(w), c(c), next(u->e) {}
} _pool[MAXM << 1], *_cur = _pool;
void addEdgeD(int  u, int v, int w, int c) {
    ND[u].e = new (_cur++) EdgeD(&ND[u], &ND[v], w, c);
    ND[v].e = new (_cur++) EdgeD(&ND[v], &ND[u], w, c);
}
namespace Dijkstra {
    struct HeapNode {
        NodeD *u;
        int dist;
        bool operator<(const HeapNode &another) const {
            return dist > another.dist;
        }
    };
    void solve(NodeD *s, int id, int n) {
        for (int i = 1; i <= n; i++) {
            ND[i].dist[id] = INT_MAX;
            ND[i].vis[id] = false;
        }
        std::priority_queue<HeapNode> q;
        s->dist[id] = 0;
        q.push((HeapNode) {s, 0});
        while (!q.empty()) {
            NodeD *u = q.top().u;
            q.pop();
            if (u->vis[id]) continue;
            u->vis[id] = true;
            for (EdgeD *e = u->e; e; e = e->next) {
                if (e->v->dist[id] > u->dist[id] + e->w) {
                    e->v->dist[id] = u->dist[id] + e->w;
                    q.push((HeapNode) {e->v, e->v->dist[id]});
                }
            }
        }
    }
}
struct Edge;
struct Node {
    Edge *e, *curr;
    int level;
} N[MAXN];
struct Edge {
    Node *u, *v;
    Edge *next, *rev;
    int cap, flow;
    Edge(Node *u, Node *v, int cap) : u(u), v(v), cap(cap), flow(0), next(u->e) {}
};
void addEdge(int u, int v, int cap) {
    N[u].e = new Edge(&N[u], &N[v], cap);
    N[v].e = new Edge(&N[v], &N[u], 0);
    N[u].e->rev = N[v].e;
    N[v].e->rev = N[u].e;
}
namespace Dinic {
    bool makeLevelGraph(Node *s, Node *t, int n) {
        for (int i = 1; i <= n; i++) N[i].level = 0;
        std::queue<Node *> q;
        q.push(s);
        s->level = 1;
        while (!q.empty()) {
            Node *u = q.front();
            q.pop();
            for (Edge *e = u->e; e; e = e->next) {
                if (e->cap > e->flow && e->v->level == 0) {
                    e->v->level = u->level + 1;
                    if (e->v == t) return true;
                    q.push(e->v);
                }
            }
        }
        return false;
    }
    int findPath(Node *s, Node *t, int limit = INT_MAX) {
        if (s == t) return limit;
        for (Edge *&e = s->curr; e; e = e->next) {
            if (e->cap > e->flow && e->v->level == s->level + 1) {
                int flow = findPath(e->v, t, std::min(limit, e->cap - e->flow));
                if (flow > 0) {
                    e->flow += flow;
                    e->rev->flow -= flow;
                    return flow;
                }
            }
        }
        return 0;
    }
    int solve(int s, int t, int n) {
        int res = 0;
        while (makeLevelGraph(&N[s], &N[t], n)) {
            for (int i = 1; i <= n; i++) N[i].curr = N[i].e;
            int flow;
            while ((flow = findPath(&N[s], &N[t])) > 0) res += flow;
        }
        return res;
    }
}
void build(int dist) {
    for (EdgeD *e = _pool; e != _cur; e++) {
        if (e->u->dist[0] + e->v->dist[1] + e->w == dist) {
            addEdge(e->u->id, e->v->id, e->c);
        }
    }
}
int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++) {
        int u, v, w, c;
        scanf("%d %d %d %d", &u, &v, &w, &c);
        addEdgeD(u, v, w, c);
    }
    Dijkstra::solve(&ND[1], 0, n);
    printf("%d\n", ND[n].dist[0]);
    Dijkstra::solve(&ND[n], 1, n);
    for (int i = 1; i <= n; i++) ND[i].id = i;
    build(ND[n].dist[0]);
    printf("%d\n", Dinic::solve(1, n, n));
    return 0;
}