HDU 3046 Pleasant sheep and big big wolf（SAP，Dinic模板）

题目地址
题意：狼要吃羊，1的位置是羊的位置，2的位置是狼的位置，0是你可以放栅栏的位置，如果羊被栅栏围到了则狼就吃不掉羊了。问要用的栅栏最少是多少？
思路：让狼与源点连，羊与汇点连，可以立栅栏的位置与周围连上一条容量为1的边，这样就转化为最小割的问题了。只要源点和汇点不连通了就隔离开来了。因为是模板题，我用了两种方法写。
Dinic：

#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <iomanip>
#define N 50000
#define M 1000000
#define LL __int64
#define inf 0x3f3f3f3f
#define lson l,mid,ans<<1
#define rson mid+1,r,ans<<1|1
#define getMid (l+r)>>1
#define movel ans<<1
#define mover ans<<1|1
using namespace std;
const LL mod = 1000000007;
int head[N], level[N];
int n, m, cnt;
struct node {
    int to;
    int cap;//剩余流量
    int next;
}edge[2 * M];
int mapp[210][210];
int dir[4][2] = { { 1,0 },{ 0,1 },{ -1,0 },{ 0,-1 } };
bool check(int x, int y) {// 判断是否越界
    if (x >= 0 && x < n && y >= 0 && y < m)
        return true;
    return false;
}
struct Dinic {
    void init() {
        memset(head, -1, sizeof(head));
        cnt = 0;
    }
    void add(int u, int v, int cap) {//有向图
        edge[cnt].to = v, edge[cnt].cap = cap, edge[cnt].next = head[u], head[u] = cnt++;
        edge[cnt].to = u, edge[cnt].cap = 0, edge[cnt].next = head[v], head[v] = cnt++;//反向边
    }
    bool bfs(int s, int t) {//建立分层图
        memset(level, -1, sizeof(level));
        queue<int>q;
        level[s] = 0;//源点的层次最高
        q.push(s);
        while (!q.empty()) {
            int u = q.front();
            q.pop();
            for (int i = head[u]; i != -1; i = edge[i].next) {
                int v = edge[i].to;
                if (edge[i].cap > 0 && level[v] < 0) {
                    level[v] = level[u] + 1;
                    q.push(v);
                    if (v == t) {
                        return true;
                    }
                }
            }
        }
        return false;
    }
    int dfs(int u, int t, int num) {//找增广路
        if (u == t || num == 0) {//找到了汇点返回当前的最小值，在这条路径上分别减去最小值
            return num;
        }
        int ans = num;
        for (int i = head[u]; i != -1; i = edge[i].next) {
            int v = edge[i].to;
            if (edge[i].cap > 0 && level[u] < level[v]) {
                int d = dfs(v, t, min(num, edge[i].cap));
                ans -= d;
                edge[i].cap -= d;
                edge[i ^ 1].cap += d;//反向边加值
                if (ans == 0)return num;
            }
        }
        return num - ans;
    }
    int dinic(int s, int t) {//源点和汇点
        int sum = 0, num;
        while (bfs(s, t)) {
            sum += dfs(s, t, inf);
        }
        return sum;
    }
}DC;
int main() {
    cin.sync_with_stdio(false);
    int Case = 1;
    while (cin >> n >> m) {
        DC.init();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                cin >> mapp[i][j];
            }
        }
        int s = n*m;
        int t = n*m + 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int pos = i*m + j;
                if (mapp[i][j] == 1) {
                    DC.add(pos, t, inf);
                }
                else if (mapp[i][j] == 2) {
                    DC.add(s, pos, inf);
                }
                for (int k = 0; k < 4; k++) {
                    int x = i + dir[k][0];
                    int y = j + dir[k][1];
                    if (check(x, y)) {
                        DC.add(pos, x*m + y, 1);
                    }
                }
            }
        }
        cout << "Case " << Case++ << ":" << endl;
        cout << DC.dinic(s, t) << endl;
    }
    return 0;
}

SAP：

#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <iomanip>
#define N 40010
#define M 800010
#define LL __int64
#define inf 0x3f3f3f3f
#define lson l,mid,ans<<1
#define rson mid+1,r,ans<<1|1
#define getMid (l+r)>>1
#define movel ans<<1
#define mover ans<<1|1
using namespace std;
const LL mod = 1000000007;
int mapp[210][210];
int head[N];
int n, m, cnt;
struct node {
    int to;
    int cap;//剩余流量
    int next;
}edge[2 * M];//因为加了正反两条边，所以要乘二
bool vis[N];
int len[N];
int gap[N];
int pre[N];
int curedge[N];
int dir[4][2] = { { 1,0 },{ 0,1 },{ -1,0 },{ 0,-1 } };
bool check(int x, int y) {// 判断是否越界
    if (x >= 0 && x < n && y >= 0 && y < m)
        return true;
    return false;
}
struct Sap {
    void init() {
        memset(head, -1, sizeof(head));
        cnt = 0;
    }
    void add(int u, int v, int cap) {//有向图
        edge[cnt].to = v, edge[cnt].cap = cap, edge[cnt].next = head[u], head[u] = cnt++;
        edge[cnt].to = u, edge[cnt].cap = 0, edge[cnt].next = head[v], head[v] = cnt++;//反向边
    }
    int max_flow(int s, int t, int n){//n为总点数个数
        int cur_flow = 0, flow_ans = 0, i, u, neck, tmp;
        memset(len, 0, sizeof(len));
        memset(gap, 0, sizeof(gap));
        memset(pre, -1, sizeof(pre));
        for (i = 0; i <= n; i++)
            curedge[i] = head[i];
        gap[0] = n + 1;
        u = s;
        while (len[s] < n + 1) {
            if (u == t) {
                cur_flow = inf;
                for (i = s; i != t; i = edge[curedge[i]].to) {
                    if (cur_flow > edge[curedge[i]].cap)
                        cur_flow = edge[curedge[i]].cap, neck = i;
                }
                for (i = s; i != t; i = edge[curedge[i]].to)
                {
                    tmp = curedge[i];
                    edge[tmp].cap -= cur_flow;
                    edge[tmp ^ 1].cap += cur_flow;
                }
                flow_ans += cur_flow;
                u = neck;
            }
            for (i = curedge[u]; i != -1; i = edge[i].next)
                if (edge[i].cap&&len[u] == len[edge[i].to] + 1)
                    break;

            if (i != -1) {
                curedge[u] = i;
                pre[edge[i].to] = u;
                u = edge[i].to;
            }
            else {
                if (0 == --gap[len[u]])
                    break;
                curedge[u] = head[u];
                for (tmp = n + 5, i = head[u]; i != -1; i = edge[i].next)
                    if (edge[i].cap)
                        tmp = min(tmp, len[edge[i].to]);
                len[u] = tmp + 1;
                ++gap[len[u]];
                if (u != s)
                    u = pre[u];
            }
        }
        return flow_ans;
    }
}sap;
int main() {
    cin.sync_with_stdio(false);
    int Case = 1;
    while (cin >> n >> m) {
        sap.init();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                cin >> mapp[i][j];
            }
        }
        int s = n*m;
        int t = n*m + 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int pos = i*m + j;
                if (mapp[i][j] == 1) {
                    sap.add(pos, t, inf);
                }
                else if (mapp[i][j] == 2) {
                    sap.add(s, pos, inf);
                }
                for (int k = 0; k < 4; k++) {
                    int x = i + dir[k][0];
                    int y = j + dir[k][1];
                    if (check(x, y)) {
                        sap.add(pos, x*m + y, 1);
                    }
                }
            }
        }
        cout << "Case " << Case++ << ":" << endl;
        cout<< sap.max_flow(s, t, t) << endl;
    }
    return 0;
}