【UOJ #107】【APIO 2013】ROBOTS

http://uoj.ac/problem/107
设(f(l,r,i,j))表示([l,r])中的机器人聚集到((i,j))需要花的最小操作数。
(f(l,r,i,j)=minleft{egin{array}{rcl} f(l,k,i,j)+f(k+1,r,i,j)\ f(l,r,i',j'),(i',j') ightarrow(i,j)end{array} ight.)
斯坦纳树的模型，对于每个([l,r])的状态，处理完第一行后再跑一遍双队列广搜处理第二行。
用基数排序优化，时间复杂度(O(n^3+n^2hw))。

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

const int N = 503;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

struct data {
	int x, y;
	data (int _x = 0, int _y = 0) : x(_x), y(_y) {}
} to[N][N][4];

bool vis[N][N][4];
char s[N][N];
int f[10][10][N][N], n, w, h;

data work(int x, int y, int tmp) {
	if (vis[x][y][tmp]) return to[x][y][tmp];
	vis[x][y][tmp] = true;
	if (s[x][y] == 'A') (++tmp) &= 3;
	if (s[x][y] == 'C') (--tmp) &= 3;
	if (s[x + dx[tmp]][y + dy[tmp]] == 'x')
		return data(x, y);
	else
		return to[x + dx[tmp]][y + dy[tmp]][tmp] = work(x + dx[tmp], y + dy[tmp], tmp);
}

int tot = 0, l, r, id[N * N];
data q1[N * N], q2[N * N], tt;

template <typename T> void check_min(T &a, T b) {if (b < a) a = b;}
template <typename T> void check_max(T &a, T b) {if (b > a) a = b;}

bool cmp(data A, data B) {return f[l][r][A.x][A.y] < f[l][r][B.x][B.y];}

int c[N * N * 20];

void BFS() {
//	stable_sort(q1 + 1, q1 + tot + 1, cmp);
	int mx = 0, num;
	for (int i = 1; i <= tot; ++i) if ((num = f[l][r][q1[i].x][q1[i].y]) != 1010580540) check_max(mx, num);
	memset(c, 0, sizeof(int) * (mx + 2));
	for (int i = 1; i <= tot; ++i) {
		num = f[l][r][q1[i].x][q1[i].y];
		if (num != 1010580540) ++c[num];
		else ++c[mx + 1];
	}
	for (int i = 1; i <= mx + 1; ++i) c[i] += c[i - 1];
	for (int i = tot; i >= 1; --i) {
		num = f[l][r][q1[i].x][q1[i].y];
		if (num == 1010580540) num = mx + 1;
		id[c[num]--] = i;
	}
	int head = 1, tail = 0, tmp = 1, x, y, x2, y2;
	while (head <= tail || tmp <= tot) {
		x = q2[head].x; y = q2[head].y;
		x2 = q1[id[tmp]].x; y2 = q1[id[tmp]].y;
		if (head <= tail && tmp <= tot && f[l][r][x][y] < f[l][r][x2][y2] || tmp > tot)	++head;
		else ++tmp, x = x2, y = y2;
		for (int d = 0; d < 4; ++d) {
			tt = to[x][y][d];
			if (tt.x == 0) continue;
			if (f[l][r][tt.x][tt.y] > f[l][r][x][y] + 1) {
				f[l][r][tt.x][tt.y] = f[l][r][x][y] + 1;
				q2[++tail] = tt;
			}
		}
	}
}

int main() {
	scanf("%d%d%d", &n, &w, &h);
	
	memset(f, 60, sizeof(f));
	
	for (int i = 1; i <= h; ++i)
		scanf("%s", s[i] + 1);
	for (int i = 1; i <= h; ++i) s[i][0] = s[i][w + 1] = 'x';
	for (int i = 1; i <= w; ++i) s[0][i] = s[h + 1][i] = 'x';
	for (int i = 1 ; i <= h; ++i)
		for (int j = 1; j <= w; ++j)
			if (s[i][j] != 'x')
				for (int dt = 0; dt < 4; ++dt)
					to[i][j][dt] = work(i, j, dt);
	
	for (int i = 1; i <= h; ++i)
		for (int j = 1; j <= w; ++j) q1[++tot] = data(i, j);
	
	bool flag;
	for (int i = n; i >= 1; --i) {
		flag = false;
		for (int ii = 1; ii <= h; ++ii) {
			for (int jj = 1; jj <= w; ++jj)
				if (s[ii][jj] == '0' + i) {
					f[i][i][ii][jj] = 0;
					flag = true;
					break;
				}
			if (flag) break;
		}
		l = r = i; BFS();
		for (int j = i + 1; j <= n; ++j) {
			for (int k = i; k < j; ++k)
				for (int ii = 1; ii <= h; ++ii)
					for (int jj = 1; jj <= w; ++jj)
						check_min(f[i][j][ii][jj], f[i][k][ii][jj] + f[k + 1][j][ii][jj]);
			l = i; r = j; BFS();
		}
	}
	
	int ans = 0x7fffffff;
	for (int i = 1; i <= h; ++i)
		for (int j = 1; j <= w; ++j)
			check_min(ans, f[1][n][i][j]);
	printf("%d
", ans == 1010580540 ? -1 : ans);
	return 0;
}