BZOJ2595（状压dp）

要点

• $f[i][j][k]$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;
}
}

posted @ 2019-05-02 22:59  AlphaWA  阅读(81)  评论(0编辑  收藏  举报