2014
11-05

# Bomberman – Just Search!

Bomberman has been a very popular game ever since it was released. As you can see above, the game is played in an N*M rectangular room. Bomberman can go around the room and place bombs. Bombs explode in 4 directions with radius r. To finish a stage, bomberman has to defeat all the foes with his bombs and find an exit behind one of the walls.
Since time is limited, bomberman has to do this job quite efficiently. Now he has successfully defeated all the foes, and is searching for the exit. It’s really troublesome to destroy the walls one by one, so he’s asking for your help to calculate the minimal number of bombs he has to place in order to destroy all the walls, thus he can surely find the exit.

The input contains several cases. Each case begins with two integers: N and M(4 <= N, M <= 15). N lines follow, each contains M characters, describing the room. A ‘*’ means a concrete wall which can never be destroyed, a ‘#’ is an ordinary wall that can be destroyed by a single bomb, a ‘.’ is an empty space where bombs can only be placed. There’re at most 30 ordinary walls. The borders of the room is always surrounded by concrete walls, as you can see from the samples. You may assume that the explosion radius r is infinite, so that an explosion can reach as far as possible until it meets a wall and destroys it if it’s an ordinary one. Proceed until the end of file.

The input contains several cases. Each case begins with two integers: N and M(4 <= N, M <= 15). N lines follow, each contains M characters, describing the room. A ‘*’ means a concrete wall which can never be destroyed, a ‘#’ is an ordinary wall that can be destroyed by a single bomb, a ‘.’ is an empty space where bombs can only be placed. There’re at most 30 ordinary walls. The borders of the room is always surrounded by concrete walls, as you can see from the samples. You may assume that the explosion radius r is infinite, so that an explosion can reach as far as possible until it meets a wall and destroys it if it’s an ordinary one. Proceed until the end of file.

9 11
***********
*#.#...#.#*
*.*.*.*.*.*
*.........*
*.*.*.*.*.*
*....#....*
*.*.*.*.*.*
*....#....*
***********
3 13
*************
*..##...##..*
*************

3
3

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int inf = (-1u>>1);
#define N 15*15
#define M N
struct node{
int r,c;
node *L,*R,*U,*D;
};
node DD[N*4+5], row[N + 5], col[M+5], head;
int cnt, size[M+5];
int n, m;
char s[20][20];
int tag[20][20];
int ans;
int rcnt, ccnt;

inline void init(int r, int c){
cnt = 0;
for (int i = c - 1; i >= 0; i--){
col[i].c = i;
col[i].r = r;
col[i].L->R = col[i].R->L = &col[i];
col[i].U = col[i].D = &col[i];
size[i] = 0;
}
for (int i = r - 1; i >= 0; i--){
row[i].r = i;
row[i].c = c;
row[i].U->D = row[i].D->U = &row[i];
row[i].L = row[i].R = &row[i];
}
}

inline void delLR(node *p){
p->L->R = p->R;
p->R->L = p->L;
}
inline void delUD(node *p){
p->U->D = p->D;
p->D->U = p->U;
}
inline void recLR(node *p){
p->L->R = p->R->L = p;
}
inline void recUD(node *p){
p->U->D = p->D->U = p;
}
inline void add(int r, int c){
node *p = &DD[cnt++];
p->c = c;
p->r = r;
p->U = &col[c];
p->D = col[c].D;
p->U->D = p->D->U = p;
p->L = &row[r];
p->R = row[r].R;
p->L->R = p->R->L = p;
size[c]++;
}

void cover(node *p){
if (p->c == ccnt)
return;
node *q;
for (q = p->D; q != p; q = q->D){
delLR(q);
}
}

void resume(node *p){
if (p->c == ccnt)
return;
node *q;
for (q = p->D; q != p; q = q->D){
recLR(q);
}
}

int h(){
bool vis[N] = {false};
node *p, *q, *pi;
int nu = 0;
if (!vis[p->c]){
vis[p->c] = true;
nu++;
for (q = col[p->c].D; q != &col[p->c]; q = q->D)
for (pi = q->R; pi != q; pi = pi->R)
vis[pi->c] = true;
}
}
return nu;
}

void DLX(int k){
node *p;
ans = k;
return;
}
if (k + h() >= ans)return;

int MIN = inf, c = 1;
if (size[p->c] < MIN){
MIN = size[p->c];
c = p->c;
}
}

for (p = col[c].D; p != (&col[c]); p = p->D){
node *q;
cover(p);
for (q = p->L; q != p; q = q->L){
cover(q);
}
DLX(k + 1);
for (q = p->R; q != p; q = q->R){
resume(q);
}
resume(p);
}
}

int main(){
//freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
while (scanf("%d %d", &n, &m) == 2){
rcnt = 0, ccnt = 0;
for (int i = 0; i < n; i++){
scanf("%s", s[i]);
for (int j = 1; s[i][j]; j++){
if (s[i][j] == '#')
tag[i][j] = ccnt++;
}
}
init((n - 2 )*(m - 2), ccnt);
for (int i = 1; i < n - 1; i++){
for (int j = 1; j < m - 1; j++){
int x, y;
if (s[i][j] == '.'){
x = i, y = j - 1;
while (s[x][y] == '.')y--;

x = i - 1, y = j;
while (s[x][y] == '.')x--;

x = i, y = j + 1;
while (s[x][y] == '.')y++;

x = i + 1, y = j;
while (s[x][y] == '.')x++;

rcnt++;
}
}
}
ans = inf;
DLX(0);
printf("%d\n", ans);
}
return 0;
}