• [BZOJ 1412][ZJOI 2009] 狼和羊的故事


    题目大意

    有一个 (n times m) 的网格,每一个格子上是羊、狼、空地中的一种,羊和狼可以走上空地。现要在格子边上建立围栏,求把狼羊分离的最少围栏数。

    (1 leqslant n, ; m leqslant 100)

    题目链接

    BZOJ 1412

    CodeVS 2351

    题解

    最小割。

    从源点向羊/狼连一条容量无限的边,从狼/羊向汇点连一条容量无限的边。考虑相邻的两格,若是一狼一羊,则连一条容量为 (1) 的边(分割狼羊),若至少有一方为空地,也连一条容量为 (1) 的边(狼羊会走上空地)。

    代码

    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    11
    12
    13
    14
    15
    16
    17
    18
    19
    20
    21
    22
    23
    24
    25
    26
    27
    28
    29
    30
    31
    32
    33
    34
    35
    36
    37
    38
    39
    40
    41
    42
    43
    44
    45
    46
    47
    48
    49
    50
    51
    52
    53
    54
    55
    56
    57
    58
    59
    60
    61
    62
    63
    64
    65
    66
    67
    68
    69
    70
    71
    72
    73
    74
    75
    76
    77
    78
    79
    80
    81
    82
    83
    84
    85
    86
    87
    88
    89
    90
    91
    92
    93
    94
    95
    96
    97
    98

    #include <climits>
    #include <queue>
    #include <algorithm>
    const int MAXN = 105;
    struct ;
    struct Node {
    Edge *e, *curr;
    int level;
    } N[MAXN * MAXN];
    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;
    }
    struct 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 = e->next) {
    大专栏  [BZOJ 1412][ZJOI 2009] 狼和羊的故事 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 operator()(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;
    }
    } dinic;
    int n, m;
    int getID(int x, int y) {
    return (x - 1) * m + y;
    }
    bool valid(int x, int y) {
    return (x > 0) && (y > 0) && (x <= n) && (y <= m);
    }
    int main() {
    scanf("%d %d", &n, &m);
    const int s = 0, t = n * m + 1;
    static int mat[MAXN][MAXN];
    for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) {
    scanf("%d", &mat[i][j]);
    if (mat[i][j] == 2) addEdge(s, getID(i, j), INT_MAX);
    if (mat[i][j] == 1) addEdge(getID(i, j), t, INT_MAX);
    }
    static int d[4][2] = {
    {0, 1},
    {0, -1},
    {1, 0},
    {-1, 0}
    };
    for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) {
    for (int k = 0; k < 4; k++) {
    int x = i + d[k][0], y = j + d[k][1];
    if (!valid(x, y)) continue;
    if (mat[i][j] != mat[x][y] || (mat[i][j] == mat[x][y] && mat[x][y] == 0))
    addEdge(getID(i, j), getID(x, y), 1);
    }
    }
    printf("%dn", dinic(s, t, t + 1));
    return 0;
    }
  • 相关阅读:
    UVA
    题目1008:最短路径问题
    怎样使用安卓变声电话软件实现边变声边通话
    Map接口及其子类
    oracle slient静默安装并配置数据库及仅安装数据库不配置数据库shell
    oracle db_unnqiue_name db_name sid_name instance_name service_name
    oracle 配置监听
    oralce dubugs
    sqlplus conn远程连接
    centos 修改shm
  • 原文地址:https://www.cnblogs.com/lijianming180/p/12099757.html
Copyright © 2020-2023  润新知