HDU3046 Pleasant sheep and big big wolf 网络流，最小割

第一道网络流的应用题，题中要求出至少要多少个围墙才能把狼阻挡在羊群活动的范围外，解决该题的方法就是以狼为源点，羊会汇点，求最小割即为最少的围墙数。出源点和汇点外其余两个网格之间的权值都是1，表示有一头狼能够通过网格走到羊群中去。

递归写的sap一直超时，最后还是递归的Dinic过了，无语了。不会非递归伤不起啊。

代码如下：

#include <cstring>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
#define RE(x) (x)^1
#define INF 0x3fffffff
#define MAXN 205
using namespace std;

int N, M, G[MAXN][MAXN], dis[MAXN*MAXN];

int dir[2][2] = {0, 1, 1, 0};

int idx, head[MAXN*MAXN];

const int sink = MAXN*MAXN-1, source = MAXN*MAXN-2;

struct Edge
{
    int v, cap, next;
}e[MAXN*MAXN*4];

inline void init()
{ 
    idx = -1;
    memset(head, 0xff, sizeof (head));
}

inline int to(int x, int y)
{
    return (x-1)*M+(y-1);
}

inline void insert(int a, int b, int c)
{
    ++idx;
    e[idx].v = b, e[idx].cap = c;
    e[idx].next = head[a], head[a] = idx;
}

inline bool judge(int x, int y)
{
    if (x < 1 || x > N || y < 1 || y > M) {
        return false;
    }
    return true;
}

inline void getint(int &t)
{
    char c;
    while (c = getchar(), c < '0' || c > '9') ;
    t = c - '0';
    while (c = getchar(), c >= '0' && c <= '9') {
        t = t * 10 + c - '0';
    }
}

inline void check(int x, int y)
{
    int xx, yy;
    if (G[x][y] == 1) {
        insert(to(x, y), sink, INF);
        insert(sink, to(x, y), 0); 
    }
    else if (G[x][y] == 2) {
        insert(source, to(x, y), INF);
        insert(to(x, y), source, 0); 
    }  
    for (int i = 0; i < 2; ++i) {
        xx = x + dir[i][0], yy = y + dir[i][1];
        if (judge(xx, yy)) {
            insert(to(x, y), to(xx, yy), 1);
            insert(to(xx, yy), to(x, y), 1);
        }
    }
}

bool bfs()
{
    int pos;
    queue<int>q;
    memset(dis, 0xff, sizeof (dis));
    dis[source] = 0;
    q.push(source);
    while (!q.empty()) {
        pos = q.front();
        q.pop();
        for (int i = head[pos]; i != -1; i = e[i].next) {
            if (dis[e[i].v] == -1 && e[i].cap > 0) {
                dis[e[i].v] = dis[pos]+1;
                q.push(e[i].v);
            }
        }
    }
    return dis[sink] != -1;
}

int dfs(int u, int flow)
{
    if (u == sink) {
        return flow;
    }
    int tf = 0, sf;
    for (int i = head[u]; i != -1; i = e[i].next) {
        if (dis[u]+1 == dis[e[i].v] && e[i].cap > 0 && (sf = dfs(e[i].v, min(flow-tf, e[i].cap)))) {
            e[i].cap -= sf, e[RE(i)].cap += sf;
            tf += sf;
            if (tf == flow) {
                return flow;
            }
        }
    }
    if (!tf) {
        dis[u] = -1;
    }
    return tf;
}

int dinic()
{
    int ans = 0;
    while (bfs()) {
        ans += dfs(source, INF);
    }
    return ans;
}

int main()
{   
    int ca = 0;
    while (scanf("%d %d", &N, &M) == 2) {
        init();
        for (int i = 1; i <= N; ++i) {
            for (int j = 1; j <= M; ++j) {
                scanf("%d", &G[i][j]);
            //    getint(G[i][j]);
            }
        }
        for (int i = 1; i <= N; ++i) {
            for (int j = 1; j <= M; ++j) {
                check(i, j);
            }
        }
        printf("Case %d:\n%d\n", ++ca, dinic());
    }
    return 0;
}