HDU - 3046 Pleasant sheep and big big wolf(最小割)

题目大意：给出一张地图，地图上面的0表示空地，1表示羊，2表示狼，现在要求你建围栏，使得所有的羊都不会被狼攻击，问至少需要建多少个围栏

解题思路：最小割，狼和羊之间的联系就是s–>u….->v–>t(u表示狼，v表示羊)，我们要做的就是断开这样的线路，且付出的代价最小，这就是最小割了
狼和源点相连接，容量为INF，羊和汇点相连接，容量为INF(刚开始弄成1了，1的话，就表示1头羊只能被一头狼攻击，这明显是错的)，两个格子之间连边，表示围栏，容量为1

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>using namespace std;const int MAXNODE = 40010;const int MAXEDGE = 400010;typedef int Type;const Type INF = 0x3f3f3f3f;struct Edge{    int u, v, next;    Type cap, flow;    Edge() {}    Edge(int u, int v, Type cap, Type flow, int next) : u(u), v(v), cap(cap), flow(flow), next(next){}};struct Dinic{    int n, m, s, t;    Edge edges[MAXEDGE];    int head[MAXNODE];    int cur[MAXNODE];    bool vis[MAXNODE];    Type d[MAXNODE];    vector<int> cut;    void init(int n) {        this->n = n;        memset(head, -1, sizeof(head));        m = 0;    }    void AddEdge(int u, int v, Type cap) {        edges[m] = Edge(u, v, cap, 0, head[u]);        head[u] = m++;        edges[m] = Edge(v, u, 0, 0, head[v]);        head[v] = m++;    }     bool BFS() {        memset(vis, 0, sizeof(vis));        queue<int> Q;        Q.push(s);        d[s] = 0;        vis[s] = 1;        while (!Q.empty()) {            int u = Q.front(); Q.pop();            for (int i = head[u]; ~i; i = edges[i].next) {                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 = edges[i].next) {            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] = head[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);        }    }}dinic;int n, m, cas = 1;int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};void solve() {    int source = 0, sink = n * m + 1;    dinic.init(sink + 1);     int t;    for (int i = 0; i < n; i++)        for (int j = 1; j <= m; j++) {            scanf("%d", &t);            if (t == 1) dinic.AddEdge(i * m + j, sink, INF);            if (t == 2) dinic.AddEdge(source, i * m + j, INF);        }    int tx, ty;    for (int i = 0; i < n; i++)        for (int j = 1; j <= m; j++)            for (int k = 0; k < 4; k++) {                tx = i + dir[k][0];                ty = j + dir[k][1];                if (tx < 0 || tx >= n || ty < 1 || ty > m) continue;                dinic.AddEdge(i * m + j, tx * m + ty, 1);            }    printf("Case %d:\n%d\n", cas++, dinic.Maxflow(source, sink));}int main() {    while (scanf("%d%d", &n, &m) != EOF) solve();    return 0;}