【BZOJ 2595】【WC 2008】游览计划

http://www.lydsy.com/JudgeOnline/problem.php?id=2595
斯坦纳树的例题诶。。。我怎么做了好长时间_(:з」∠)_
首先这是一棵树。
状压表示状态,\(f(i,j,s)\)表示连通的景点的状态为s,i和j为树根的最小值。
转移时先在当前状态s上枚举s的子集t,用子集来转移\(f(i,j,s)=min\{f(i,j,t)+f(i,j,\complement_st)-a(i,j)\},t\varsubsetneqq s,t\neq\varnothing\)
顺便把所有可以用来更新的状态加入队列,然后在当前枚举这层枚举的s内做spfa,求出其他点的f值。
因为最优解一定可以通过两个没有重叠方块的连通块合并,所以不用担心求出的解会有方块被算多次。
时间复杂度\(O(2^kn^2m^2)\)

#include<cstdio>
#include<cstring>
#include<algorithm>
#define mk(x, y, z) ((x) + (y) * 100 + (z) * 10000)
using namespace std;

int a[13][13], f[13][13][1 << 10], pre[13][13][1 << 10], n, m, q[1000003], head, tail, inf, tot = 0, ans = 0x7fffffff;
bool inq[10003], mark[13][13];

const int dx[4] = {0, -1, 0, 1};
const int dy[4] = {1, 0, -1, 0};

void spfa(int s) {
	int u, x, y, tx, ty, t;
	while (head != tail) {
		++head; if (head == 1000003) head = 0;
		u = q[head]; inq[u] = false;
		x = u % 100; y = u / 100;
		for (int i = 0; i < 4; ++i) {
			tx = x + dx[i]; ty = y + dy[i];
			if (tx < 1 || tx > n || ty < 1 || ty > m) continue;
			
			t = f[x][y][s] + a[tx][ty];
			if (t < f[tx][ty][s]) {
				f[tx][ty][s] = t;
				pre[tx][ty][s] = mk(x, y, s);
				u = mk(tx, ty, 0);
				if (!inq[u]) {
					inq[u] = true;
					++tail; if (tail == 1000003) tail = 0;
					q[tail] = u;
				}
			}
		}
	}
}

void dfs(int x, int y, int s) {
	mark[x][y] = true;
	int num = pre[x][y][s]; if (num == 0) return;
	if (num / 10000 == s)
		dfs(num % 100, num / 100 % 100, s);
	else {
		dfs(x, y, num / 10000);
		dfs(x, y, s ^ (num / 10000));
	}
}

void ouit(int totnum) {
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			if (f[i][j][totnum] == ans) {
				dfs(i, j, totnum);
				return;
			}
}

int main() {
	scanf("%d%d", &n, &m);
	memset(f, 60, sizeof(f)); inf = f[0][0][0];
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j) {
			scanf("%d", &a[i][j]);
			if (!a[i][j]) {
				f[i][j][1 << tot] = 0;
				++tot;
			}
		}
	
	int totnum = (1 << tot) - 1, now;
	for (int s = 1; s <= totnum; ++s) {
		head = tail = 0;
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= m; ++j) {
				for (int t = s & (s - 1); t; t = s & (t - 1)) {
					now = f[i][j][t] + f[i][j][s ^ t] - a[i][j];
					if (now < f[i][j][s]) {
						f[i][j][s] = now;
						pre[i][j][s] = mk(i, j, t);
					}
				}
				if (f[i][j][s] != inf)
					inq[q[++tail] = mk(i, j, 0)] = true;
		}
		spfa(s);
	}
	
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			ans = min(ans, f[i][j][totnum]);
	printf("%d\n", ans);
	
	ouit(totnum);
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j)
			if (mark[i][j])	putchar(a[i][j] ? 'o' : 'x');
			else putchar('_');
		puts("");
	}
	return 0;
}
posted @ 2016-12-15 21:02  abclzr  阅读(213)  评论(0编辑  收藏  举报