PAT (Top Level) Practise1019Separate the Animals (35)

1019. Separate the Animals (35)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

LIU, Rujia

There are some animals in a zoo which can be described as a grid with N rows and M columns. Your task is to place some obstacles so that no pairs of animals can reach each other.

Two animals can reach each other if and only if their cells are 4-connected. For example, in Figure 1, the central blue cell can be reached by the four red cells, and cannot be reached by the other four white cells.

Figure 1

What is more, you must put obstacles in exactly K cells, which are 4-connected and form exactly H holes. Here a hole is defined as a 4-connected part with finitely many open cells while the zoo is placed in an infinite open grid. For example, there are 2 holes (the green and the yellow areas) in Figure 2.

Figure 2

For the following grid with two animals:

Figure 3

If K = 8 and H = 1, one way to separate them is the following:

Figure 4

Figure 5 is illegal because it contains no hole.

Figure 5

Figure 6 is also illegal because the obstacles are not 4-connected.

Figure 6

Given some animals, you are supposed to count the number of different solutions.

Input Specification:

Each input file contains one test case. For each case, the first line gives four integers: N, M, K, H (2 <= N, M <= 6; 1 <= K <= 12; 0 <= H <= 2). All the numbers are separated by spaces.

Then N lines follow, each contains M characters, which are either "." or "O", representing an open cell or an animal, respectively. There will be at least 2 animals.

Output Specification:

For each case, print a single line containing a single integer: the number of solutions.

Sample Input:

3 5 8 1...O..O........

Sample Output:

8

爆搜加剪枝，我写的有点繁琐了，不知道有没有巧妙的办法

#include<map> #include<set>#include<ctime>  #include<cmath>      #include<queue>   #include<string>  #include<vector>  #include<cstdio>      #include<cstring>    #include<iostream>  #include<algorithm>      #include<functional>  using namespace std;#define ms(x,y) memset(x,y,sizeof(x))      #define rep(i,j,k) for(int i=j;i<=k;i++)      #define per(i,j,k) for(int i=j;i>=k;i--)      #define loop(i,j,k) for (int i=j;i!=-1;i=k[i])      #define inone(x) scanf("%d",&x)      #define intwo(x,y) scanf("%d%d",&x,&y)      #define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)    #define infou(x,y,z,p) scanf("%d%d%d%d",&x,&y,&z,&p)   #define lson x<<1,l,mid  #define rson x<<1|1,mid+1,r  #define mp(i,j) make_pair(i,j)  #define ff first  #define ss second  typedef int LL;typedef pair<LL, LL> pii;const int low(int x) { return x&-x; }const double eps = 1e-6;const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int N = 15;int n, m, t, h;int fa[N*N][N*N], ga[N*N][N*N], sa[N*N][N*N], a[N][N], ans, vis[N][N];int ha[N*N][N*N], hole[N], b[N*N];char s[N][N];int dir[4][2] = { 0,1,0,-1,1,0,-1,0 };int get(int fa[], int x) {  return fa[x] == x ? x : fa[x] = get(fa, fa[x]);}void dfs(int now, int cnt, int con) {  if (cnt > t) return;  if (con - 1 > t - cnt) return;  if (t - cnt > n*m - now + 1) return;  if (now > n * m) {    if (cnt < t || con != 1 || hole[now - 1] != h) return;    ans++;     return;  }  int x = (now - 1) / m + 1, y = (now - 1) % m + 1;  rep(i, 1, n*m) {    fa[now][i] = fa[now - 1][i];    ga[now][i] = ga[now - 1][i];    sa[now][i] = sa[now - 1][i];    ha[now][i] = ha[now - 1][i];  }  hole[now] = hole[now - 1] + 1;  if (con == 1 && cnt == t) {  }  else if (now > m) {    int sz = 0;    rep(i, 1, m) {      int x = (now - i - 1) / m + 1, y = (now - i - 1) % m + 1;      if (!a[x][y]) continue;      b[sz++] = get(fa[now], now - i);    }    sort(b, b + sz);    sz = unique(b, b + sz) - b;    if (sz < con) {      return;    }  }  if (y > 1 && !a[x][y - 1]) {    int fx = get(ha[now], now), fy = get(ha[now], now - 1);    if (fx != fy) {      if (!fx || !fy) ha[now][fx + fy] = 0;      else  ha[now][fx] = fy;      hole[now]--;    }  }  if (x > 1 && !a[x - 1][y]) {    int fx = get(ha[now], now), fy = get(ha[now], now - m);    if (fx != fy) {      if (!fx || !fy) ha[now][fx + fy] = 0;      else  ha[now][fx] = fy;      hole[now]--;    }  }  if (y==1 || y==m || x == 1 || x == n) {    int fx = get(ha[now], now);    if (fx) {      ha[now][fx] = 0; hole[now]--;    }  }  int flag = 1;  if (y > 1 && !a[x][y - 1]) {    int fx = get(ga[now], now), fy = get(ga[now], now - 1);    if (fx != fy) {       ga[now][fx] = fy;       if ((sa[now][fy] += sa[now][fx]) > 1) flag = 0;    }  }  if (x > 1 && !a[x - 1][y]) {    int fx = get(ga[now], now), fy = get(ga[now], now - m);    if (fx != fy) {      ga[now][fx] = fy;      if ((sa[now][fy] += sa[now][fx]) > 1) flag = 0;    }  }  if (flag) dfs(now + 1, cnt, con);  if (s[x][y] == 'O') return;  rep(i, 1, n*m) {    fa[now][i] = fa[now - 1][i];    ga[now][i] = ga[now - 1][i];    sa[now][i] = sa[now - 1][i];    ha[now][i] = ha[now - 1][i];  }  hole[now] = hole[now - 1];  a[x][y] = 1;  con++;  if (y > 1 && a[x][y-1]) {    int fx = get(fa[now], now), fy = get(fa[now], now - 1);    if (fx != fy) { fa[now][fx] = fy; con--; }  }  if (x > 1 && a[x-1][y]) {    int fx = get(fa[now], now), fy = get(fa[now], now - m);    if (fx != fy) { fa[now][fx] = fy; con--; }  }  dfs(now + 1, cnt + 1, con);  a[x][y] = 0;}int main() {  scanf("%d%d%d%d", &n, &m, &t, &h);  rep(i, 1, n) scanf("%s", s[i] + 1);  rep(i, 1, n) rep(j, 1, m) {    fa[0][i*m - m + j] = i*m - m + j;    ga[0][i*m - m + j] = i*m - m + j;    ha[0][i*m - m + j] = i*m - m + j;    sa[0][i*m - m + j] = s[i][j] == 'O';    hole[0] = 0;  }  ans = 0;  dfs(1, 0, 0);  printf("%d\n", ans);  return 0;}