[洛谷P2774]方格取数问题

题目大意：给你一个$n\times m$的方格，要求你从中选择一些数，其中没有相邻两个数，使得最后和最大

题解：网络流，最小割，发现相邻的两个点不可以同时选择，进行黑白染色，原点向黑点连一条容量为点权的边，白点向汇点连一条容量为点权的边，黑点向周围一圈的白点连容量为$inf$的边，总权值减去跑出来的最小割就是答案。

卡点：无

C++ Code：

#include <algorithm>
#include <cstdio>
#include <cctype>
namespace __IO {
	namespace R {
		int x, ch;
		inline int read() {
			ch = getchar();
			while (isspace(ch)) ch = getchar();
			for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
			return x;
		}
	}
}
using __IO::R::read;

#define maxn 100010
#define maxm (maxn << 3)
const int inf = 0x3f3f3f3f;

int n, m, sum;
namespace Network_Flow {
	int st, ed, MF;
	int head[maxn], cnt = 1;
	struct Edge {
		int to, nxt, w;
	} e[maxm << 1];
	inline void addedge(int a, int b, int c) {
		e[++cnt] = (Edge) {b, head[a], c}; head[a] = cnt;
		e[++cnt] = (Edge) {a, head[b], 0}; head[b] = cnt;
	}

	int GAP[maxn], d[maxn], lst[maxn];
	int q[maxn], h, t;
	void init() {
		GAP[d[ed] = 1] = 1;
		for (int i = st; i <= ed; i++) lst[i] = head[i];
		q[h = t = 0] = ed;
		while (h <= t) {
			int u = q[h++];
			for (int i = head[u]; i; i = e[i].nxt) {
				int v = e[i].to;
				if (!d[v]) {
					d[v] = d[u] + 1;
					GAP[d[v]]++;
					q[++t] = v;
				}
			}
		}
	}
	int dfs(int u, int low) {
		if (!low || u == ed) return low;
		int w, res = 0;
		for (int &i = lst[u]; i; i = e[i].nxt) if (e[i].w) {
			int v = e[i].to;
			if (d[u] == d[v] + 1) {
				w = dfs(v, std::min(low, e[i].w));
				res += w, low -= w;
				e[i].w -= w, e[i ^ 1].w += w;
				if (!low) return res;
			}
		}
		if (!(--GAP[d[u]])) d[st] = ed + 1;
		++GAP[++d[u]], lst[u] = head[u];
		return res;
	}
	void ISAP(int __st, int __ed) {
		st = __st, ed = __ed;
		init();
		while (d[st] <= ed) MF += dfs(st, inf);
	}
}
using Network_Flow::addedge;

inline int get(int x, int y) {return (x - 1) * m + y;}
const int D[2][4] = {{0, 1, 0, -1}, {1, 0, -1, 0}};
inline bool over_range(int x, int y) {
	return x < 1 || x > n || y < 1 || y > m;
}
int main() {
	n = read(), m = read();
	int st = 0, ed = n * m + 1;
	for (int i = 1, x, pos; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			sum += x = read(), pos = get(i, j);
			if (i + j & 1) {
				addedge(st, pos, x);
				for (int k = 0; k < 4; k++) {
					int __x = i + D[0][k], __y = j + D[1][k];
					if (!over_range(__x, __y)) addedge(pos, get(__x, __y), inf);
				}
			} else addedge(pos, ed, x);
		}
	}
	Network_Flow::ISAP(st, ed);
	printf("%d\n", sum - Network_Flow::MF);
	return 0;
}