[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; }