HDU 3046 Pleasant sheep and big big wolf(最小割)
HDU 3046 Pleasant sheep and big big wolf
题意:一个n * m平面上,1是羊。2是狼,问最少要多少围墙才干把狼所有围住,每有到达羊的路径
思路:有羊和狼。要分成两个集合互不可达。显然的最小割。建图源点连狼,容量无穷,羊连汇点,容量无穷。然后相邻格子连边。容量为1
代码:
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std; const int MAXNODE = 40005;
const int MAXEDGE = 500005; typedef int Type;
const Type INF = 0x3f3f3f3f; struct Edge {
int u, v;
Type cap, flow;
Edge() {}
Edge(int u, int v, Type cap, Type flow) {
this->u = u;
this->v = v;
this->cap = cap;
this->flow = flow;
}
}; struct Dinic {
int n, m, s, t;
Edge edges[MAXEDGE];
int first[MAXNODE];
int next[MAXEDGE];
bool vis[MAXNODE];
Type d[MAXNODE];
int cur[MAXNODE];
vector<int> cut; void init(int n) {
this->n = n;
memset(first, -1, sizeof(first));
m = 0;
}
void add_Edge(int u, int v, Type cap) {
edges[m] = Edge(u, v, cap, 0);
next[m] = first[u];
first[u] = m++;
edges[m] = Edge(v, u, 0, 0);
next[m] = first[v];
first[v] = m++;
} bool bfs() {
memset(vis, false, sizeof(vis));
queue<int> Q;
Q.push(s);
d[s] = 0;
vis[s] = true;
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = first[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (!vis[e.v] && e.cap > e.flow) {
vis[e.v] = true;
d[e.v] = d[u] + 1;
Q.push(e.v);
}
}
}
return vis[t];
} Type dfs(int u, Type a) {
if (u == t || a == 0) return a;
Type flow = 0, f;
for (int &i = cur[u]; i != -1; i = next[i]) {
Edge& e = edges[i];
if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[i^1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
} Type Maxflow(int s, int t) {
this->s = s; this->t = t;
Type flow = 0;
while (bfs()) {
for (int i = 0; i < n; i++)
cur[i] = first[i];
flow += dfs(s, INF);
}
return flow;
} void MinCut() {
cut.clear();
for (int i = 0; i < m; i += 2) {
if (vis[edges[i].u] && !vis[edges[i].v])
cut.push_back(i);
}
}
} gao; const int N = 205;
const int d[4][2] = {0, 1, 0, -1, 1, 0, -1, 0}; int n, m, g[N][N]; int main() {
int cas = 0;
while (~scanf("%d%d", &n, &m)) {
gao.init(n * m + 2);
int s = n * m, t = n * m + 1;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
scanf("%d", &g[i][j]);
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (g[i][j] == 2) gao.add_Edge(s, i * m + j, INF);
if (g[i][j] == 1) gao.add_Edge(i * m + j, t, INF);
for (int k = 0; k < 4; k++) {
int x = i + d[k][0];
int y = j + d[k][1];
if (x < 0 || x >= n || y < 0 || y >= m) continue;
gao.add_Edge(i * m + j, x * m + y, 1);
}
}
}
printf("Case %d:\n", ++cas);
printf("%d\n", gao.Maxflow(s, t));
}
return 0;
}
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