昂神的解题报告：http://sd-invol.github.io/2014/10/22/Anshan-2014-G/

我来对他的话进行翻译就好了… 

之所以看出最小割  是因为每个位置有两种方案  这样形成二分图后  我们要进行决策  最小割也就变成了进行决策所要丢掉的最小价值

之所以根据每个位置的二进制表示中1的个数来决定该位置两种决策放在左边还是右边  是为了避免二分图中同一个集合里的两个点连边

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<bitset>
using namespace std;
#define N 300
#define inf 50000000

int T, S, n, m, tot, cas, ans;
int head[N * 2], f[N], g[N], Less[N], Lid[N], More[N], Mid[N], dis[N * 2], qu[N
		* 2];
struct edge {
	int v, w, next;
} ed[N * N * 100];

int cut(int x) {
	int i;
	for (i = 0; x; x -= (x & (-x)), i++)
		;
	return i;
}

void add(int u, int v, int w) {
	ed[tot].v = v;
	ed[tot].w = w;
	ed[tot].next = head[u];
	head[u] = tot++;
	ed[tot].v = u;
	ed[tot].w = 0;
	ed[tot].next = head[v];
	head[v] = tot++;
}

bool bfs() {
	int l, r, u, v, i;
	memset(dis, -1, sizeof(dis));
	dis[S] = 0;
	qu[0] = S;
	l = 0;
	r = 1;
	while (l < r) {
		u = qu[l++];
		for (i = head[u]; ~i; i = ed[i].next) {
			v = ed[i].v;
			if (dis[v] < 0 && ed[i].w > 0) {
				dis[v] = dis[u] + 1;
				if (v == T)
					return true;
				qu[r++] = v;
			}
		}
	}
	return false;
}

int dfs(int u, int nowflow) {
	if (u == T)
		return nowflow;
	int i, v, tmp, res = 0;
	for (i = head[u]; ~i; i = ed[i].next) {
		v = ed[i].v;
		if (dis[v] == dis[u] + 1 && ed[i].w > 0) {
			tmp = dfs(v, min(nowflow, ed[i].w));
			nowflow -= tmp;
			ed[i].w -= tmp;
			ed[i ^ 1].w += tmp;
			res += tmp;
			if (!nowflow)
				break;
		}
	}
	if (!nowflow)
		dis[u] = -1;
	return res;
}

int main() {
	scanf("%d", &cas);
	while (cas--) {
		scanf("%d%d", &n, &m);
		n = 1 << n;
		m = 1 << m;
		tot = 0;
		memset(head, -1, sizeof(head));
		ans = 0;
		memset(Less, -1, sizeof(Less));
		memset(More, -1, sizeof(More));
		S = 0;
		T = n * 2 + 1;
		for (int i = 0; i < n; i++)
			scanf("%d", &f[i]);
		for (int i = 0; i < n; i++)
			scanf("%d", &g[i]);
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				int k;
				scanf("%d", &k);
				k += 1024;
				if (j < f[i]) {
					if (Less[i] < k) {
						Less[i] = k;
						Lid[i] = j;
					}
				} else {
					if (More[i] < k) {
						More[i] = k;
						Mid[i] = j;
					}
				}
			}
		}
		for (int i = 0; i < n; i++) {
			int aa = cut(i);
			if (aa & 1) {
				add(S, i + 1, Less[i]);
				add(i + 1 + n, T, More[i]);
			} else {
				add(S, i + 1, More[i]);
				add(i + 1 + n, T, Less[i]);
			}
			add(i + 1, i + 1 + n, inf);
			for (int j = i + 1; j < n; j++) {
				if (cut(i ^ j) == 1) {
					if (aa & 1)
						add(i + 1, j + 1 + n, g[i] ^ g[j]);
					else
						add(j + 1, i + 1 + n, g[i] ^ g[j]);
				}
			}
		}
		while (bfs())
			ans += dfs(S, inf);
		for (int i = 0; i < n; i++) {
			if (i)
				printf(" ");
			int aa = cut(i);
			if (aa & 1) {
				if (dis[i + 1] != -1)
					printf("%d", Lid[i]);
				else
					printf("%d", Mid[i]);
			} else {
				if (dis[i + 1] != -1)
					printf("%d", Mid[i]);
				else
					printf("%d", Lid[i]);
			}
		}
		printf("\n");
	}
	return 0;
}

HDU 5076 Memory