pku 1753 Filp Game for循环+剪枝 > bfs

呵呵, 我用的是循环遍历, 每层加剪枝, 相当于bfs吧!

不过比bfs用的空间少的多.

Rank:93

6223070

Memory:164K

Time:0MS

C++

1950B

2009-12-09 09:45:29

 

 #include <cstdio> using namespace std; bool map[4][4]; inline bool get_bit(int num, int i) { if(i == -1 || i == 4) return 0; else return (num >> i)&1; } //-1: not ok //0: all 0 //1: all 1 inline int check(int i, int j, int k, int line)//check line i, j next line, k pre line { int m; bool temp[4] = {0}; for(m = 3; m >= 0; --m) { if(get_bit(i, m+1)^ get_bit(i, m-1)^ get_bit(i, m)^ get_bit(j, m)^ get_bit(k, m)) temp[m] = !map[line][m]; else temp[m] = map[line][m]; } for(m = 1; m < 4; ++m) if(temp[0] != temp[m]) return -1; if(temp[0]) return 1; else return 0; } inline int get_bit_count(int n) { int cnt = 0; for(int i = 0; i < 4; ++i) if(get_bit(n, i)) cnt++; return cnt; } int main() { int i, j, k, l; bool flagOccurWhite = false; bool flagOccurBlack = false; char ch; for(i = 0; i < 4; ++i) { for(j = 0; j < 4; ++j) { scanf("%c", &ch); if(ch == 'w') { map[i][j] = true; flagOccurWhite = true; } else { map[i][j] = false; flagOccurBlack = true; } } getchar(); } if(flagOccurBlack == false || flagOccurWhite == false) { printf("0/n"); return 0; } int min = 17; int v1, v2, v3, v4, temp; for(i = 0; i < 16; ++i) for(j = 0; j < 16; ++j) { v1 = check(i, j, 0, 0); if(v1 == -1) continue; for(k = 0; k < 16; ++k) { v2 = check(j, k, i, 1); if(v2 == -1 || v2 != v1) continue; for(l = 0; l < 16; ++l) { v3 = check(k, l, j, 2); if(v3 == -1 || v3 != v2) continue; v4 = check(l, 0, k, 3); if(v4 == -1 || v4 != v3) continue; temp = get_bit_count(i)+get_bit_count(j) +get_bit_count(k)+get_bit_count(l); if(temp < min) min = temp; } } } if(min < 17) printf("%d/n", min); else printf("Impossible/n"); return 0; }

 

 

code2:

 

#include <iostream> #include <queue> using namespace std; int n; bool visited[65536]; int step[65536]; inline void change(int& num, int i) { num ^= (1<<i); if(i%4 != 0) num ^= (1<<(i-1)); if(i%4 != 3) num ^= (1<<(i+1)); if(i/4 != 0) num ^= (1<<(i-4)); if(i/4 != 3) num ^= (1<<(i+4)); } int bfs() { memset(visited, 0, sizeof(visited)); memset(step, 0, sizeof(step)); queue<int> q; q.push(n); visited[n] = true; while(!q.empty()) { int v = q.front(); visited[v] = true; q.pop(); for(int i = 0; i < 16; ++i) { int temp = v; change(temp, i); if(temp == 0 || temp == 65535) return step[v]+1; if(!visited[temp]) { q.push(temp); visited[temp] = true; step[temp] = step[v]+1; } } } return -1; } int main() { char ch; for(int i = 0; i < 4; ++i) for(int j = 0; j < 4; ++j) { cin >> ch; if(ch == 'w') n += 1; n <<= 1; } n >>= 1; //move only 15 times if(n == 0 || n == 65535) { printf("0/n"); return 0; } int ret = bfs(); if(ret != -1) printf("%d/n", ret); else printf("Impossible/n"); return 0; }

 总结一下:

这种用16位的二进制数表示4*4方格所有状态的方法也很巧妙.

 具体说来,用16位二进制数表示一个4*4方格的当前状态,
 通过对这个16位二进制数的任意一位进行变化后,进入队列.

 不过效率显然不如用for循环+剪枝来的快, 因为先对一行的状态
 进行遍历, 剪枝会多一些.

 可以发现, bfs的精髓是: 采用二进制数做状态的转移的量. bfs不过是
 保存状态和寻找他的下一状态.