BZOJ2595（状压dp）

要点

• 设\(f[i][j][k]\)为经过点\((i,j)\)且包含点集\(k\)的最小代价，其中k是指景点集合的枚举。
• 考虑有两种情况：1.点\((i,j)\)作为关键点连接了两个子集时\(f[i][j][k]\)可以得到最小，有\(f[i][j][k]=f[i][j][k_1]+f[i][j][k_2]-a[i][j],\ k_1|k_2=k\)；2.点\((i,j)\)作为主干道发展出来的枝叶时\(f[i][j][k]\)可以得到最小，那么它是它的邻居发展过来的，spfa跑一下即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

const int inf = 0x3f3f3f3f;
const int xx[] = {0, 0, -1, 1};
const int yy[] = {1, -1, 0, 0};

int N, M, K, a[11][11];
int f[11][11][1 << 11], pre[11][11][1 << 11][3], vis[11][11];
queue< pair<int, int> > Q;

void dfs(int i, int j, int U) {
	if (!U)	return;
	vis[i][j] = 1;
	int a = pre[i][j][U][0], b = pre[i][j][U][1], c = pre[i][j][U][2];
	dfs(a, b, c);
	if (i == a && j == b)	dfs(a, b, c ^ U);
}

void output() {
	for (int i = 1; i <= N; i++) {
		for (int j = 1; j <= M; j++)
			if (!a[i][j])	putchar('x');
			else if (vis[i][j])	putchar('o');
			else	putchar('_');
		puts("");
	}
}

int main() {
	scanf("%d %d", &N, &M);
	for (int i = 1; i <= N; i++)
		for (int j = 1; j <= M; j++) {
			for (int k = 0; k < (1 << 10); k++) {
				f[i][j][k] = inf;
			}
			scanf("%d", &a[i][j]);
			if (!a[i][j]) {
				f[i][j][1 << (K++)] = 0;
			}
		}
	for (int U = 1; U < (1 << K); U++) {
		memset(vis, 0, sizeof vis);
		//作为枝干的转移
		for (int i = 1; i <= N; i++)
			for (int j = 1; j <= M; j++) {
				for (int k = U & (U - 1); k; k = U & (k - 1)) {//枚举二进制真子集
					int tmp = f[i][j][k] + f[i][j][k ^ U] - a[i][j];
					if (tmp < f[i][j][U]) {
						f[i][j][U] = tmp;
						pre[i][j][U][0] = i;
						pre[i][j][U][1] = j;
						pre[i][j][U][2] = k;
					}
				}
				if (f[i][j][U] < inf)	Q.push({i, j}), vis[i][j] = 1;
			}
		//作为树枝的转移
		while (Q.size()) {
			int x = Q.front().first, y = Q.front().second; Q.pop();
			vis[x][y] = 0;
			for (int t = 0; t < 4; t++) {
				int nx = x + xx[t], ny = y + yy[t];
				if (nx < 1 || nx > N || ny < 1 || ny > M)	continue;
				if (f[nx][ny][U] > f[x][y][U] + a[nx][ny]) {
					f[nx][ny][U] = f[x][y][U] + a[nx][ny];
					pre[nx][ny][U][0] = x;
					pre[nx][ny][U][1] = y;
					pre[nx][ny][U][2] = U;
					if (!vis[nx][ny])	Q.push({nx, ny}), vis[nx][ny] = 1;
				}
			}
		}
	}
	for (int i = 1; i <= N; i++)
		for (int j = 1; j <= M; j++)
			if (!a[i][j]) {
				memset(vis, 0, sizeof vis);
				dfs(i, j, (1 << K) - 1);
				printf("%d\n", f[i][j][(1 << K) - 1]);
				output();
				return 0;
			}
}