【LOJ6007】【网络流24题】方格取数（最小割）

Description

233

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Solution

将图进行黑白染色，将其看作一个二分图，从超级源点向白点连边，流量为其权值；从黑点向超级汇点连边，流量也为权值。黑、白点之间如果相邻则连流量为INF的边。然后用总权值减去最小割即为答案。

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Code

/************************************************
 * Au: Hany01
 * Date: May 30th, 2018
 * Prob: [LOJ6007][网络流24题] 方格取数
 * Email: [email protected]
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
    register int _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 105 * 105, maxm = 105 * 105 * 10;

int n, m, S, T, beg[maxn], v[maxm], nex[maxm], f[maxm], e = 1, gap[maxn], d[maxn];

#define id(x,y) ((x - 1) * m + y)

inline void add(int uu, int vv, int ff) { v[++ e] = vv, f[e] = ff, nex[e] = beg[uu], beg[uu] = e; }
inline void Add(int uu, int vv, int ff) { add(uu, vv, ff), add(vv, uu, 0); }

int sap(int u, int flow)
{
    if (u == T) return flow;
    int res = flow, tmp;
    for (register int i = beg[u]; i; i = nex[i])
        if (f[i] && d[v[i]] + 1 == d[u])
        {
            tmp = sap(v[i], min(f[i], res));
            f[i] -= tmp, f[i ^ 1] += tmp;
            if (!(res -= tmp)) return flow;
        }
    if (!-- gap[d[u] ++]) d[T] = T;
    ++ gap[d[u]];
    return flow - res;
}

int main()
{
#ifdef hany01
    File("loj6007");
#endif

    static LL Ans = 0, Sum = 0;
    static int tmp;

    n = read(), m = read(), S = n * m + 1, T = S + 1;
    For(i, 1, n) For(j, 1, m) {
        Sum += (tmp = read());
        if ((i + j) & 1) {
            Add(S, id(i, j), tmp);
            if (i > 1) Add(id(i, j), id(i - 1, j), INF);
            if (j > 1) Add(id(i, j), id(i, j - 1), INF);
            if (i < n) Add(id(i, j), id(i + 1, j), INF);
            if (j < m) Add(id(i, j), id(i, j + 1), INF);
        } else Add(id(i, j), T, tmp);
    }

    for (gap[0] = T; d[T] < T; ) Ans += sap(S, INF);
    printf("%lld\n", Sum - Ans);

    return 0;
}
//东飞伯劳西飞燕，黄姑织女时相见。
//    -- 萧衍《东飞伯劳歌》