Description
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。
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》