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  • [BZOJ 3144][HNOI 2013] 切糕

    题目大意

    切糕是 (p times q times r) 的长方体,每个点有一个违和感 (v_{x, y, z})。先要水平切开切糕(即对于每个纵轴,切面与其有且只有一个交点),要求水平上相邻两点的切面高度差小于等于 (D),求切面违和感和的最小值。

    (1 leqslant p, ; q, ; r leqslant 40)

    (0 leqslant v leqslant 1,000)

    题目链接

    BZOJ 3144

    CodeVS 2997

    题解

    最小割。

    用边连接相邻两个高度的的点,边 ((x, y, z - 1) rightarrow (x, y, z)) 容量为 (v_{x, y, z}),由源点发散出边连接第一层的每个点,最后一层的点收缩在汇点,这是没有(D)的限制是的答案。连接所有形如 ((x, y, z) rightarrow (x, y, z - D)) 的边,这样,当水平相邻的两个点切面差大于 (D) 时,最小割的图会由这样的边连在一起而没有被隔开。

    代码

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    #include <climits>
    #include <queue>
    #include <algorithm>
    const int MAXN = 40;
    struct ;
    struct Node {
    Edge *e, *curr;
    int level;
    } N[MAXN * MAXN * MAXN + 2];
    struct {
    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 = 0; 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 大专栏  [BZOJ 3144][HNOI 2013] 切糕= 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 = 0; i < n; i++) N[i].curr = N[i].e;
    int flow;
    while ((flow = findPath(&N[s], &N[t])) > 0) res += flow;
    }
    return res;
    }
    }
    int p, q, r;
    int getID(int x, int y, int z) {
    if (z == 0) return 0;
    return (z - 1) * p * q + (x - 1) * q + y;
    }
    bool valid(int x, int y) {
    return (x > 0) && (y > 0) && (x <= p) && (y <= q);
    }
    int main() {
    int D;
    scanf("%d %d %d %d", &p, &q, &r, &D);
    static int v[MAXN + 1][MAXN + 1][MAXN + 1];
    for (int k = 1; k <= r; k++) for (int i = 1; i <= p; i++) for (int j = 1; j <= q; j++) scanf("%d", &v[i][j][k]);
    const int s = 0, t = p * q * r + 1;
    const int d[4][2] = {
    {0, 1},
    {0, -1},
    {1, 0},
    {-1, 0}
    };
    for (int i = 1; i <= p; i++) for (int j = 1; j <= q; j++) {
    for (int k = 1; k <= r; k++) {
    addEdge(getID(i, j, k - 1), getID(i, j, k), v[i][j][k]);
    if (k > D) for (int l = 0; l < 4; l++) {
    int x = i + d[l][0], y = j + d[l][1];
    if (valid(x, y)) addEdge(getID(i, j, k), getID(x, y, k - D), INT_MAX);
    }
    }
    addEdge(getID(i, j, r), t, INT_MAX);
    }
    printf("%dn", Dinic::solve(s, t, t + 1));
    return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lijianming180/p/12037853.html
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