ACM题目————网格动物

Lattice animal is a set of connected sites on a lattice. Lattice animals on a square lattice are especially popular subject of study and are also known as polyominoes. Polyomino is usually represented as a set of sidewise connected squares. Polyomino with n squares is called n-polyomino. In this problem you are to find a number of distinct free n-polyominoes that fit into rectangle w×h. Free polyominoes can be rotated and flipped over, so that their rotations and mirror images are considered to be the same. For example, there are 5 different pentominoes (5-polyominoes) that fit into 2×4 rectangle and 3 different octominoes (8-polyominoes) that fit into 3×3 rectangle.

Input 

The input file contains several test cases, one per line. This line consists of 3 integer numbers n, w, and h ( 1n10, 1w, hn).

Output 

For each one of the test cases, write to the output file a single line with a integer number -- the number of distinct free n-polyominoes that fit into rectangle w×h.

Sample Input 

5 1 4
5 2 4
5 3 4
5 5 5
8 3 3

Sample Output 

0
5
11
12
3

输入n，w，h （1<=n<=10,1<=w,h<=n），求能放在w*h网格里的不同的n连块的个数(注意,平移，旋转，翻转后相同的算作同一种)。例如，2*4里的5连块有5种(第一行)，而3*3里的8连块有以下3种(第二行)。

 

 

难点：

 

1.以每个格子来扩展。先枚举1连块，在对1连块的每个格子的4个方向进行扩展，枚举2连块，依次类推。

 

2.将n连块表示成n个格子的集合，将所有的n连块又表示成集合，判重任务交给set.

 

3.判重时要将n连块进行8个方向的旋转，并且每个n连块需要规范化(左下角的格子在(0,0)).

 

4.得到n连块后判断是否能放进w*h的网格中，由于n连块已经规范化，得到n连块的格子最大x,y坐标，即能盛下该n连块的长和宽。

 

 

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#include <cstdio> #include <set> #include <algorithm> using namespace std; #define maxn 10 // 代表一个网格节点 typedef struct cell {     int x, y;   //网格节点的坐标       // 构造函数     cell(int x, int y)     {         this->x = x;         this->y = y;     }       bool operator < (const struct cell& a) const     {         return x < a.x || (x == a.x && y < a.y);     } }cell;   // 一个Polyomino就是一堆cell的集合 typedef set<cell> poly;   // poly_set[i]代表有i个cell的poly集合 set<poly> poly_set[maxn+1];   // answer[n][w][h]的答案 int answer[maxn+1][maxn+1][maxn+1];   void gen_poly(); void check_poly(const poly& this_p, cell& this_c); poly normalize(poly& p); poly rotate(poly& p); poly flip(poly& p);     int main() {     // 生成所有poly     gen_poly();   //  printf("here\n");     int n, w, h;     while(scanf("%d %d %d", &n, &w, &h) == 3)     {         printf("%d\n", answer[n][w][h]);     }     return 0; }   int dic_x[4] = {-1,0,1,0}; int dic_y[4] = {0,1,0,-1};   // 生成所有poly void gen_poly() {     for(int i = 1; i <= maxn; i++)         poly_set[i] = set<poly>();       // 先生成有1个cell的poly     poly p1;     p1.insert(cell(0,0));     poly_set[1].insert(p1);       // 分别根据有i-1个cell的poly集合来生成有i个cell的poly集合     for(int i = 2; i <= maxn; i++)     {         // 对每个poly中的每个cell尝试在不同的四个方向增加一个cell         for(set<poly>::iterator p = poly_set[i-1].begin(); p != poly_set[i-1].end(); p++)         {             for(poly::const_iterator q = p->begin(); q != p->end(); q++)             {                 for(int j = 0; j < 4; j++)                 {                     cell new_c(q->x+dic_x[j], q->y+dic_y[j]); //                  cell new_c;                     if(p->find(new_c) == p->end())                     {                         // 检查形成的这个poly是否存在，如果不存在就加入                         check_poly(*p, new_c);                     }                   }             }         }     }       // 对所有n,w,h生成答案     for(int i = 1; i <= maxn; i++)     {         for(int w = 1; w <= i; w++)         {             for(int h = 1; h <= i; h++)             {                 int count = 0;                 for(set<poly>::iterator p = poly_set[i].begin(); p != poly_set[i].end(); p++)                         {                     int max_x = p->begin()->x, max_y = p->begin()->y;                     for(poly::iterator q = p->begin(); q != p->end(); q++)                     {                         if(max_x < q->x)                             max_x = q->x;                         if(max_y < q->y)                             max_y = q->y;                     }                       if(min(max_x, max_y) < min(w, h) && max(max_x, max_y) < max(w, h))                     {                         count++;                     }                 } /*              if(count != 0)                     printf("answer[%d][%d][%d] = %d\n", i, w, h, count); */              answer[i][w][h] = count;             }         }     } }     // 检查形成的这个poly加上这个cell是否存在，如果不存在就加入 void check_poly(const poly& this_p, cell& this_c) {     poly p = this_p;     p.insert(this_c);     // 规范化到最小点为(0,0)     p = normalize(p);       int n = p.size();     // 检查旋转的8个方向是否存在，如果不存在就加入到poly集合     for(int i = 0; i < 4; i++)     {         if(poly_set[n].find(p) != poly_set[n].end())             return;         // 对该poly向右旋转90度         p = rotate(p);     }     // 将该poly向下反转180度     p = flip(p);     for(int i = 0; i < 4; i++)         {                 if(poly_set[n].find(p) != poly_set[n].end())                         return;                 // 对该poly向右旋转90度                 p = rotate(p);         }     poly_set[n].insert(p);   }   // 规范化到最小点为(0,0) poly normalize(poly& p) {     poly this_p;     int min_x = p.begin()->x, min_y = p.begin()->y;     for(poly::iterator q = p.begin(); q != p.end(); q++)     {         if(q->x < min_x)             min_x = q->x;         if(q->y < min_y)             min_y = q->y;     }     for(poly::iterator q = p.begin(); q != p.end(); q++)         {         this_p.insert(cell(q->x-min_x,q->y-min_y));         }     return this_p; }   // 对该poly向右旋转90度 poly rotate(poly& p) {     poly this_p;     for(poly::iterator q = p.begin(); q != p.end(); q++)         {                 this_p.insert(cell(q->y,-q->x));         }         return normalize(this_p); }     // 将该poly向下反转180度 poly flip(poly& p) {     poly this_p;         for(poly::iterator q = p.begin(); q != p.end(); q++)         {                 this_p.insert(cell(q->x,-q->y));         }         return normalize(this_p); }