BZOJ 1001: [BeiJing2006]狼抓兔子

二次联通门 : BZOJ 1001: [BeiJing2006]狼抓兔子

 lqz已经废了

/*    
　　BZOJ 1001: [BeiJing2006]狼抓兔子

    最小割

*/
#include <cstdio>
#include <iostream>
#include <queue>
#define rg register
inline void read (int &n) {
    rg char c = getchar ();
    for (n = 0; !isdigit (c); c = getchar ());
    for (; isdigit (c); n = n * 10 + c - '0', c = getchar ());
}
inline int min (int a, int b) { return a < b ? a : b; }
int S, T;
namespace net {
    const int Max = 1000006;
    int _v[Max * 6], _n[Max * 6], list[Max], _f[Max * 6], EC = 1;
    int q[Max * 6];
    inline void In (int u, int v, int f) {
        _v[++ EC] = v, _n[EC] = list[u], list[u] = EC, _f[EC] = f;
        _v[++ EC] = u, _n[EC] = list[v], list[v] = EC, _f[EC] = f;
    }

    int d[Max];
    bool Bfs ()    {
        int h = 0, t = 1;
        q[h] = S; rg int i, n;
        for (i = 0; i <= T; ++ i) d[i] = -1;
        for (d[S] = 0; h < t; ++ h) {
            n = q[h];
            for (i = list[n]; i; i = _n[i])
                if (_f[i] && d[_v[i]] == -1) {
                    d[_v[i]] = d[n] + 1, q[t ++] = _v[i];
                    if (_v[i] == T) return true;
                }
        }
        return false;
    }
    
    int Flowing (int n, int f) {
        if (n == T || f == 0) return f;
        int p, r = 0; 
        for (rg int i = list[n]; i; i = _n[i])
            if (d[_v[i]] == d[n] + 1 && _f[i]) {
                p = Flowing (_v[i], min (f, _f[i]));
                if (p > 0) { 
                    r += p, f -= p, _f[i] -= p, _f[i ^ 1] += p;
                    if (f == 0) return r; 
                }
            }
        if (r != f) d[n] = -1; return r;
    }
#define INF 1000000000
    inline int Dinic (int S, int T) { 
        int Answer = 0;
        for (; Bfs (); Answer += Flowing (S, INF)); 
        return Answer;
    }
}

int main (int argc, char *argv[]) { 

    int N, M, x; read (N), read (M); rg int i, j;
    for (i = 1; i <= N; ++ i)
        for (j = 1; j < M; ++ j)
            read (x), net :: In ((i - 1) * M + j, (i - 1) * M + j + 1, x);
    for (i = 1; i < N; ++ i)
        for (j = 1; j <= M; ++ j)
            read (x), net :: In ((i - 1) * M + j, i * M + j, x);
    for (i = 1; i < N; ++ i)
        for (j = 1; j < M; ++ j)
            read (x), net :: In ((i - 1) * M + j, i * M + j + 1, x);
    S = 1, T = N * M;
    printf ("%d", net :: Dinic (S, T));
    return 0;
}